\displaystyle \textbf{Question 1: }~\text{If }\overrightarrow{a}=\widehat{i}+3\widehat{j}-2\widehat{k}\text{ and }\overrightarrow{b}=-\widehat{i}+3\widehat{k},\text{ find }\left|\overrightarrow{a}\times\overrightarrow{b}\right|.
\displaystyle \text{Answer:}
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=\widehat{i}+3\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{b}=-\widehat{i}+0\widehat{j}+3\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&3&-2\\-1&0&3\end{vmatrix}
\displaystyle =(3\cdot3-(-2)\cdot0)\widehat{i}-(1\cdot3-(-2)\cdot(-1))\widehat{j}+(1\cdot0-3\cdot(-1))\widehat{k}
\displaystyle =(9+0)\widehat{i}-(3-2)\widehat{j}+(0+3)\widehat{k}
\displaystyle =9\widehat{i}-\widehat{j}+3\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{9^2+(-1)^2+3^2}
\displaystyle =\sqrt{91}

\displaystyle \textbf{Question 2: }~\text{(i) If }\overrightarrow{a}=3\widehat{i}+4\widehat{j}\text{ and }\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k},\text{ find the value of }\left|\overrightarrow{a}\times\overrightarrow{b}\right|.
\displaystyle \text{(ii) If }\overrightarrow{a}=2\widehat{i}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k},\text{ find the magnitude of }\overrightarrow{a}\times\overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=3\widehat{i}+4\widehat{j}+0\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&4&0\\1&1&1\end{vmatrix}
\displaystyle =(4\cdot1-0\cdot1)\widehat{i}-(3\cdot1-0\cdot1)\widehat{j}+(3\cdot1-4\cdot1)\widehat{k}
\displaystyle =(4-0)\widehat{i}-(3-0)\widehat{j}+(3-4)\widehat{k}
\displaystyle =4\widehat{i}-3\widehat{j}-\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{4^2+(-3)^2+(-1)^2}
\displaystyle =\sqrt{26}
\displaystyle \text{(ii) }
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=2\widehat{i}+0\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&0&1\\1&1&1\end{vmatrix}
\displaystyle =(0\cdot1-1\cdot1)\widehat{i}-(2\cdot1-1\cdot1)\widehat{j}+(2\cdot1-0\cdot1)\widehat{k}
\displaystyle =(0-1)\widehat{i}-(2-1)\widehat{j}+(2-0)\widehat{k}
\displaystyle =-\widehat{i}-\widehat{j}+2\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(-1)^2+(-1)^2+2^2}
\displaystyle =\sqrt{6}

\displaystyle \textbf{Question 3: }~\text{(i) Find a unit vector perpendicular to both the vectors } \\ 4\widehat{i}-\widehat{j}+3\widehat{k}\text{ and }-2\widehat{i}+\widehat{j}-2\widehat{k}.
\displaystyle \text{(ii) Find a unit vector perpendicular to the plane containing the vectors } \\ \overrightarrow{a}=2\widehat{i}+\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=4\widehat{i}-\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b}=-2\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&-1&3\\-2&1&-2\end{vmatrix}
\displaystyle =( (-1)\cdot(-2)-3\cdot1 )\widehat{i}-( 4\cdot(-2)-3\cdot(-2) )\widehat{j}+( 4\cdot1-(-1)\cdot(-2) )\widehat{k}
\displaystyle =(2-3)\widehat{i}-(-8+6)\widehat{j}+(4-2)\widehat{k}
\displaystyle =-\widehat{i}+2\widehat{j}+2\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(-1)^2+2^2+2^2}
\displaystyle =\sqrt{9}=3
\displaystyle \text{Unit vector perpendicular to }\overrightarrow{a}\text{ and }\overrightarrow{b}=\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}=\frac{-\widehat{i}+2\widehat{j}+2\widehat{k}}{3}
\displaystyle \text{(ii) }
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=2\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&1\\1&2&1\end{vmatrix}
\displaystyle =(1\cdot1-1\cdot2)\widehat{i}-(2\cdot1-1\cdot1)\widehat{j}+(2\cdot2-1\cdot1)\widehat{k}
\displaystyle =(1-2)\widehat{i}-(2-1)\widehat{j}+(4-1)\widehat{k}
\displaystyle =-\widehat{i}-\widehat{j}+3\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(-1)^2+(-1)^2+3^2}
\displaystyle =\sqrt{11}
\displaystyle \text{Unit vector perpendicular to the plane containing vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}=\pm\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}
\displaystyle =\pm\frac{1}{\sqrt{11}}\left(-\widehat{i}-\widehat{j}+3\widehat{k}\right)

\displaystyle \textbf{Question 4: }~\text{Find the magnitude of vector }\overrightarrow{a}=(3\widehat{k}+4\widehat{j})\times(\widehat{i}+\widehat{j}-\widehat{k}).
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}=(0\widehat{i}+4\widehat{j}+3\widehat{k})\times(\widehat{i}+\widehat{j}-\widehat{k})
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\0&4&3\\1&1&-1\end{vmatrix}
\displaystyle =\widehat{i}(4\cdot(-1)-3\cdot1)-\widehat{j}(0\cdot(-1)-3\cdot1)+\widehat{k}(0\cdot1-4\cdot1)
\displaystyle =\widehat{i}(-4-3)-\widehat{j}(0-3)+\widehat{k}(0-4)
\displaystyle =-7\widehat{i}+3\widehat{j}-4\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}|=\sqrt{(-7)^2+3^2+(-4)^2}
\displaystyle =\sqrt{74}

\displaystyle \textbf{Question 5: }~\text{If }\overrightarrow{a}=4\widehat{i}+3\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}-2\widehat{k},\text{ then find }\left|2\overrightarrow{b}\times\overrightarrow{a}\right|.
\displaystyle \text{Answer:}
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=4\widehat{i}+3\widehat{j}+\widehat{k}
\displaystyle 2\overrightarrow{b}=2\widehat{i}+0\widehat{j}-4\widehat{k}
\displaystyle \therefore \ 2\overrightarrow{b}\times\overrightarrow{a}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&0&-4\\4&3&1\end{vmatrix}
\displaystyle =(0\cdot1-(-4)\cdot3)\widehat{i}-(2\cdot1-(-4)\cdot4)\widehat{j}+(2\cdot3-0\cdot4)\widehat{k}
\displaystyle =(0+12)\widehat{i}-(2+16)\widehat{j}+(6-0)\widehat{k}
\displaystyle =12\widehat{i}-18\widehat{j}+6\widehat{k}
\displaystyle \Rightarrow |2\overrightarrow{b}\times\overrightarrow{a}|=\sqrt{12^2+(-18)^2+6^2}
\displaystyle =\sqrt{504}

\displaystyle \textbf{Question 6: }~\text{If }\overrightarrow{a}=3\widehat{i}-\widehat{j}-2\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+3\widehat{j}+\widehat{k},\text{ find } \\ (\overrightarrow{a}+2\overrightarrow{b})\times(2\overrightarrow{a}-\overrightarrow{b}).
\displaystyle \text{Answer:}
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=3\widehat{i}-\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{b}=2\widehat{i}+3\widehat{j}+\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}+2\overrightarrow{b}=3\widehat{i}-\widehat{j}-2\widehat{k}+2(2\widehat{i}+3\widehat{j}+\widehat{k})
\displaystyle =7\widehat{i}+5\widehat{j}+0\widehat{k}
\displaystyle \therefore \ 2\overrightarrow{a}-\overrightarrow{b}=2(3\widehat{i}-\widehat{j}-2\widehat{k})-(2\widehat{i}+3\widehat{j}+\widehat{k})
\displaystyle =4\widehat{i}-5\widehat{j}-5\widehat{k}
\displaystyle (\overrightarrow{a}+2\overrightarrow{b})\times(2\overrightarrow{a}-\overrightarrow{b})=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\7&5&0\\4&-5&-5\end{vmatrix}
\displaystyle =(5\cdot(-5)-0\cdot(-5))\widehat{i}-(7\cdot(-5)-0\cdot4)\widehat{j}+(7\cdot(-5)-5\cdot4)\widehat{k}
\displaystyle =(-25+0)\widehat{i}-(-35+0)\widehat{j}+(-35-20)\widehat{k}
\displaystyle =-25\widehat{i}+35\widehat{j}-55\widehat{k}

\displaystyle \textbf{Question 7: }~\text{(i) Find a vector of magnitude }49,\text{ which is perpendicular} \\ \text{to both the vectors }2\widehat{i}+3\widehat{j}+6\widehat{k}\text{ and }3\widehat{i}-6\widehat{j}+2\widehat{k}.
\displaystyle \text{(ii) Find a vector whose length is }3\text{ and which is perpendicular to the} \\ \text{vector }\overrightarrow{a}=3\widehat{i}+\widehat{j}-4\widehat{k}\text{ and }\overrightarrow{b}=6\widehat{i}+5\widehat{j}-2\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=2\widehat{i}+3\widehat{j}+6\widehat{k}
\displaystyle \overrightarrow{b}=3\widehat{i}-6\widehat{j}+2\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&6\\3&-6&2\end{vmatrix}
\displaystyle =(3\cdot2-6\cdot(-6))\widehat{i}-(2\cdot2-6\cdot3)\widehat{j}+(2\cdot(-6)-3\cdot3)\widehat{k}
\displaystyle =(6+36)\widehat{i}-(4-18)\widehat{j}+(-12-9)\widehat{k}
\displaystyle =42\widehat{i}+14\widehat{j}-21\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{42^2+14^2+(-21)^2}
\displaystyle =\sqrt{2401}=49
\displaystyle \text{Required vector }=49\left(\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}\right)
\displaystyle =49\left(\frac{42\widehat{i}+14\widehat{j}-21\widehat{k}}{49}\right)
\displaystyle =42\widehat{i}+14\widehat{j}-21\widehat{k}
\displaystyle \text{(ii) }
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=3\widehat{i}+\widehat{j}-4\widehat{k}
\displaystyle \overrightarrow{b}=6\widehat{i}+5\widehat{j}-2\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&1&-4\\6&5&-2\end{vmatrix}
\displaystyle =(1\cdot(-2)-(-4)\cdot5)\widehat{i}-(3\cdot(-2)-(-4)\cdot6)\widehat{j}+(3\cdot5-1\cdot6)\widehat{k}
\displaystyle =(-2+20)\widehat{i}-(-6+24)\widehat{j}+(15-6)\widehat{k}
\displaystyle =18\widehat{i}-18\widehat{j}+9\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{18^2+(-18)^2+9^2}
\displaystyle =\sqrt{729}=27
\displaystyle \text{Required vector }=3\left(\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}\right)
\displaystyle =3\left(\frac{18\widehat{i}-18\widehat{j}+9\widehat{k}}{27}\right)
\displaystyle =2\widehat{i}-2\widehat{j}+\widehat{k}

\displaystyle \textbf{Question 8: }~\text{Find the area of the parallelogram determined by the vectors:}
\displaystyle \text{(i) }2\widehat{i}\text{ and }3\widehat{j}
\displaystyle \text{(ii) }2\widehat{i}+\widehat{j}+3\widehat{k}\text{ and }\widehat{i}-\widehat{j}
\displaystyle \text{(iii) }3\widehat{i}+\widehat{j}-2\widehat{k}\text{ and }\widehat{i}-3\widehat{j}+4\widehat{k}
\displaystyle \text{(iv) }\widehat{i}-3\widehat{j}+\widehat{k}\text{ and }\widehat{i}+\widehat{j}+\widehat{k}\qquad [\text{CBSE 2002C}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Let :}
\displaystyle \overrightarrow{a}=2\widehat{i}+0\widehat{j}+0\widehat{k}
\displaystyle \overrightarrow{b}=0\widehat{i}+3\widehat{j}+0\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&0&0\\0&3&0\end{vmatrix}
\displaystyle =(0\cdot0-0\cdot3)\widehat{i}-(2\cdot0-0\cdot0)\widehat{j}+(2\cdot3-0\cdot0)\widehat{k}
\displaystyle =0\widehat{i}+0\widehat{j}+6\widehat{k}
\displaystyle \text{Area of the parallelogram }=|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\sqrt{0^2+0^2+6^2}
\displaystyle =6\ \text{sq. units}
\displaystyle \text{(ii) }
\displaystyle \text{Let :}
\displaystyle \overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}-\widehat{j}+0\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&3\\1&-1&0\end{vmatrix}
\displaystyle =(1\cdot0-3\cdot(-1))\widehat{i}-(2\cdot0-3\cdot1)\widehat{j}+(2\cdot(-1)-1\cdot1)\widehat{k}
\displaystyle =(0+3)\widehat{i}-(0-3)\widehat{j}+(-2-1)\widehat{k}
\displaystyle =3\widehat{i}+3\widehat{j}-3\widehat{k}
\displaystyle \text{Area of the parallelogram }=|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\sqrt{3^2+3^2+(-3)^2}
\displaystyle =\sqrt{27}=3\sqrt{3}\ \text{sq. units}
\displaystyle \text{(iii) }
\displaystyle \text{Let :}
\displaystyle \overrightarrow{a}=3\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}-3\widehat{j}+4\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&1&-2\\1&-3&4\end{vmatrix}
\displaystyle =(1\cdot4-(-2)\cdot(-3))\widehat{i}-(3\cdot4-(-2)\cdot1)\widehat{j}+(3\cdot(-3)-1\cdot1)\widehat{k}
\displaystyle =(4-6)\widehat{i}-(12+2)\widehat{j}+(-9-1)\widehat{k}
\displaystyle =-2\widehat{i}-14\widehat{j}-10\widehat{k}
\displaystyle \text{Area of the parallelogram }=|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\sqrt{(-2)^2+(-14)^2+(-10)^2}
\displaystyle =\sqrt{300}=10\sqrt{3}\ \text{sq. units}
\displaystyle \text{(iv) }
\displaystyle \text{Let :}
\displaystyle \overrightarrow{a}=\widehat{i}-3\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-3&1\\1&1&1\end{vmatrix}
\displaystyle =( (-3)\cdot1-1\cdot1 )\widehat{i}-( 1\cdot1-1\cdot1 )\widehat{j}+( 1\cdot1-(-3)\cdot1 )\widehat{k}
\displaystyle =(-3-1)\widehat{i}-(1-1)\widehat{j}+(1+3)\widehat{k}
\displaystyle =-4\widehat{i}+0\widehat{j}+4\widehat{k}
\displaystyle \text{Area of the parallelogram }=|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\sqrt{(-4)^2+0^2+4^2}
\displaystyle =\sqrt{32}=4\sqrt{2}\ \text{sq. units}

\displaystyle \textbf{Question 9: }~\text{Find the area of the parallelogram whose diagonals are:}
\displaystyle \text{(i) }4\widehat{i}-\widehat{j}-3\widehat{k}\text{ and }-2\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \text{(ii) }2\widehat{i}+\widehat{k}\text{ and }\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \text{(iii) }3\widehat{i}+4\widehat{j}\text{ and }\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \text{(iv) }2\widehat{i}+3\widehat{j}+6\widehat{k}\text{ and }3\widehat{i}-6\widehat{j}+2\widehat{k}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Let :}
\displaystyle \overrightarrow{a}=4\widehat{i}-\widehat{j}-3\widehat{k}
\displaystyle \overrightarrow{b}=-2\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\4&-1&-3\\-2&1&-2\end{vmatrix}
\displaystyle =( (-1)\cdot(-2)-(-3)\cdot1 )\widehat{i}-( 4\cdot(-2)-(-3)\cdot(-2) )\widehat{j}+( 4\cdot1-(-1)\cdot(-2) )\widehat{k}
\displaystyle =(2+3)\widehat{i}-(-8-6)\widehat{j}+(4-2)\widehat{k}
\displaystyle =5\widehat{i}+14\widehat{j}+2\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{5^2+14^2+2^2}
\displaystyle =\sqrt{225}=15
\displaystyle \text{Area of the triangle }=\frac{1}{2}|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\frac{15}{2}\ \text{sq. units}
\displaystyle \text{(ii) }
\displaystyle \text{Let :}
\displaystyle \overrightarrow{a}=2\widehat{i}+0\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&0&1\\1&1&1\end{vmatrix}
\displaystyle =(0\cdot1-1\cdot1)\widehat{i}-(2\cdot1-1\cdot1)\widehat{j}+(2\cdot1-0\cdot1)\widehat{k}
\displaystyle =(0-1)\widehat{i}-(2-1)\widehat{j}+(2-0)\widehat{k}
\displaystyle =-\widehat{i}-\widehat{j}+2\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{(-1)^2+(-1)^2+2^2}
\displaystyle =\sqrt{6}
\displaystyle \text{Area of the parallelogram }=\frac{1}{2}|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\frac{\sqrt{6}}{2}\ \text{sq. units}
\displaystyle \text{(iii) }
\displaystyle \text{Let :}
\displaystyle \overrightarrow{a}=3\widehat{i}+4\widehat{j}+0\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&4&0\\1&1&1\end{vmatrix}
\displaystyle =(4\cdot1-0\cdot1)\widehat{i}-(3\cdot1-0\cdot1)\widehat{j}+(3\cdot1-4\cdot1)\widehat{k}
\displaystyle =(4-0)\widehat{i}-(3-0)\widehat{j}+(3-4)\widehat{k}
\displaystyle =4\widehat{i}-3\widehat{j}-\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{4^2+(-3)^2+(-1)^2}
\displaystyle =\sqrt{26}
\displaystyle \text{Area of the parallelogram }=\frac{1}{2}|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\frac{\sqrt{26}}{2}\ \text{sq. units}
\displaystyle \text{(iv) }
\displaystyle \text{Let :}
\displaystyle \overrightarrow{a}=2\widehat{i}+3\widehat{j}+6\widehat{k}
\displaystyle \overrightarrow{b}=3\widehat{i}-6\widehat{j}+2\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&6\\3&-6&2\end{vmatrix}
\displaystyle =(3\cdot2-6\cdot(-6))\widehat{i}-(2\cdot2-6\cdot3)\widehat{j}+(2\cdot(-6)-3\cdot3)\widehat{k}
\displaystyle =(6+36)\widehat{i}-(4-18)\widehat{j}+(-12-9)\widehat{k}
\displaystyle =42\widehat{i}+14\widehat{j}-21\widehat{k}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{42^2+14^2+(-21)^2}
\displaystyle =\sqrt{2401}=49
\displaystyle \text{Area of the parallelogram }=\frac{1}{2}|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\frac{49}{2}\ \text{sq. units}

\displaystyle \textbf{Question 10: }~\text{If }\overrightarrow{a}=2\widehat{i}+5\widehat{j}-7\widehat{k},\ \overrightarrow{b}=-3\widehat{i}+4\widehat{j}+\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}-2\widehat{j}-3\widehat{k},\text{ compute }(\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c}\text{ and }\overrightarrow{a}\times(\overrightarrow{b}\times\overrightarrow{c})\text{ and verify that these are not equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=2\widehat{i}+5\widehat{j}-7\widehat{k}
\displaystyle \overrightarrow{b}=-3\widehat{i}+4\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{c}=\widehat{i}-2\widehat{j}-3\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&5&-7\\-3&4&1\end{vmatrix}
\displaystyle =(5\cdot1-(-7)\cdot4)\widehat{i}-(2\cdot1-(-7)\cdot(-3))\widehat{j}+(2\cdot4-5\cdot(-3))\widehat{k}
\displaystyle =(5+28)\widehat{i}-(2-21)\widehat{j}+(8+15)\widehat{k}
\displaystyle =33\widehat{i}+19\widehat{j}+23\widehat{k}
\displaystyle \therefore (\overrightarrow{a}\times\overrightarrow{b})\times\overrightarrow{c}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\33&19&23\\1&-2&-3\end{vmatrix}
\displaystyle =(19\cdot(-3)-23\cdot(-2))\widehat{i}-(33\cdot(-3)-23\cdot1)\widehat{j}+(33\cdot(-2)-19\cdot1)\widehat{k}
\displaystyle =(-57+46)\widehat{i}-(-99-23)\widehat{j}+(-66-19)\widehat{k}
\displaystyle =-11\widehat{i}+122\widehat{j}-85\widehat{k}\qquad (1)
\displaystyle \therefore \ \overrightarrow{b}\times\overrightarrow{c}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-3&4&1\\1&-2&-3\end{vmatrix}
\displaystyle =(4\cdot(-3)-1\cdot(-2))\widehat{i}-((-3)\cdot(-3)-1\cdot1)\widehat{j}+((-3)\cdot(-2)-4\cdot1)\widehat{k}
\displaystyle =(-12+2)\widehat{i}-(9-1)\widehat{j}+(6-4)\widehat{k}
\displaystyle =-10\widehat{i}-8\widehat{j}+2\widehat{k}
\displaystyle \therefore \ \overrightarrow{a}\times(\overrightarrow{b}\times\overrightarrow{c})=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&5&-7\\-10&-8&2\end{vmatrix}
\displaystyle =(5\cdot2-(-7)\cdot(-8))\widehat{i}-(2\cdot2-(-7)\cdot(-10))\widehat{j}+(2\cdot(-8)-5\cdot(-10))\widehat{k}
\displaystyle =(10-56)\widehat{i}-(4-70)\widehat{j}+(-16+50)\widehat{k}
\displaystyle =-46\widehat{i}+66\widehat{j}+34\widehat{k}\qquad (2)
\displaystyle \text{From (1) and (2), we get }(\overrightarrow{a}\times\overrightarrow{b})\times\overrightarrow{c}\neq\overrightarrow{a}\times(\overrightarrow{b}\times\overrightarrow{c}).

\displaystyle \textbf{Question 11: }~\text{If }\lvert\overrightarrow{a}\rvert=2,\ \lvert\overrightarrow{b}\rvert=5\text{ and }\lvert\overrightarrow{a}\times\overrightarrow{b}\rvert=8,\text{ find }\overrightarrow{a}\cdot\overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{We know}
\displaystyle (\overrightarrow{a}\cdot\overrightarrow{b})^2+|\overrightarrow{a}\times\overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2
\displaystyle \Rightarrow (\overrightarrow{a}\cdot\overrightarrow{b})^2+8^2=2^2\times5^2\ (\because |\overrightarrow{a}\times\overrightarrow{b}|=8,\ |\overrightarrow{a}|=2\text{ and }|\overrightarrow{b}|=5)
\displaystyle \Rightarrow (\overrightarrow{a}\cdot\overrightarrow{b})^2+64=100
\displaystyle \Rightarrow (\overrightarrow{a}\cdot\overrightarrow{b})^2=36
\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=6

\displaystyle \textbf{Question 12: }~\text{Given }\overrightarrow{a}=\frac{1}{7}(2\widehat{i}+3\widehat{j}+6\widehat{k}),  \overrightarrow{b}=\frac{1}{7}(3\widehat{i}-6\widehat{j}+2\widehat{k}),\\ \overrightarrow{c}=\frac{1}{7}(6\widehat{i}+2\widehat{j}-3\widehat{k}),\ \widehat{i},\widehat{j},\widehat{k}\text{ being a right handed orthogonal system of unit} \\ \text{vectors in space, show that }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ is also another system.}
\displaystyle \text{Answer:}
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=\frac{1}{7}(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \overrightarrow{b}=\frac{1}{7}(3\widehat{i}-6\widehat{j}+2\widehat{k})
\displaystyle \overrightarrow{c}=\frac{1}{7}(6\widehat{i}+2\widehat{j}-3\widehat{k})
\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\left(\frac{1}{7}\right)\left(\frac{1}{7}\right)\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&6\\3&-6&2\end{vmatrix}
\displaystyle =\frac{1}{49}(42\widehat{i}+14\widehat{j}-21\widehat{k})
\displaystyle =\frac{1}{49}\left[7(6\widehat{i}+2\widehat{j}-3\widehat{k})\right]
\displaystyle =\frac{1}{7}(6\widehat{i}+2\widehat{j}-3\widehat{k})
\displaystyle =\overrightarrow{c}
\displaystyle \overrightarrow{b}\times\overrightarrow{c}=\left(\frac{1}{7}\right)\left(\frac{1}{7}\right)\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-6&2\\6&2&-3\end{vmatrix}
\displaystyle =\frac{1}{49}(14\widehat{i}+21\widehat{j}+42\widehat{k})
\displaystyle =\frac{1}{49}\left[7(2\widehat{i}+3\widehat{j}+6\widehat{k})\right]
\displaystyle =\frac{1}{7}(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle =\overrightarrow{a}
\displaystyle \overrightarrow{c}\times\overrightarrow{a}=\left(\frac{1}{7}\right)\left(\frac{1}{7}\right)\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\6&2&-3\\2&3&6\end{vmatrix}
\displaystyle =\frac{1}{49}(21\widehat{i}-42\widehat{j}+14\widehat{k})
\displaystyle =\frac{1}{49}\left[7(3\widehat{i}-6\widehat{j}+2\widehat{k})\right]
\displaystyle =\frac{1}{7}(3\widehat{i}-6\widehat{j}+2\widehat{k})
\displaystyle =\overrightarrow{b}
\displaystyle |\overrightarrow{a}|=\frac{1}{7}\sqrt{2^2+3^2+6^2}
\displaystyle =\frac{1}{7}\sqrt{4+9+36}
\displaystyle =\frac{7}{7}=1
\displaystyle |\overrightarrow{b}|=\frac{1}{7}\sqrt{3^2+(-6)^2+2^2}
\displaystyle =\frac{1}{7}\sqrt{9+36+4}
\displaystyle =\frac{7}{7}=1
\displaystyle |\overrightarrow{c}|=\frac{1}{7}\sqrt{6^2+2^2+(-3)^2}
\displaystyle =\frac{1}{7}\sqrt{36+4+9}
\displaystyle =\frac{7}{7}=1
\displaystyle \text{Thus, }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ form a right handed orthogonal system of unit vectors.}

\displaystyle \textbf{Question 13: }~\text{If }\lvert\overrightarrow{a}\rvert=13,\ \lvert\overrightarrow{b}\rvert=5\text{ and }\overrightarrow{a}\cdot\overrightarrow{b}=60,\text{ then find }\left|\overrightarrow{a}\times\overrightarrow{b}\right|.
\displaystyle \text{Answer:}
\displaystyle \text{We know}
\displaystyle (\overrightarrow{a}\cdot\overrightarrow{b})^2+|\overrightarrow{a}\times\overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2
\displaystyle \Rightarrow (60)^2+|\overrightarrow{a}\times\overrightarrow{b}|^2=13^2\times5^2\ (\because \overrightarrow{a}\cdot\overrightarrow{b}=60,\ |\overrightarrow{a}|=13\text{ and }|\overrightarrow{b}|=5)
\displaystyle \Rightarrow 3600+|\overrightarrow{a}\times\overrightarrow{b}|^2=4225
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|^2=625
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=25

\displaystyle \textbf{Question 14: }~\text{Find the angle between two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ if } \\ \left|\overrightarrow{a}\times\overrightarrow{b}\right|=\overrightarrow{a}\cdot\overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.
\displaystyle \text{Given : }|\overrightarrow{a}\times\overrightarrow{b}|=\overrightarrow{a}\cdot\overrightarrow{b}
\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\sin\theta=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta
\displaystyle \Rightarrow \sin\theta=\cos\theta
\displaystyle \Rightarrow \tan\theta=1
\displaystyle \Rightarrow \theta=\frac{\pi}{4}

\displaystyle \textbf{Question 15: }~\text{If }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}\ne \overrightarrow{0},\text{ then show that } \\ \overrightarrow{a}+\overrightarrow{c}=m\overrightarrow{b},\text{ where }m\text{ is any scalar.}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}
\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{b}=-\,\overrightarrow{c}\times\overrightarrow{b}
\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{c}\times\overrightarrow{b}=0
\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{c})\times\overrightarrow{b}=0\ \text{(using right distributive property)}
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{c}\ \text{is parallel to }\overrightarrow{b}
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{c}=m\,\overrightarrow{b},\ \text{for some scalar }m

\displaystyle \textbf{Question 16: }~\text{If }\lvert\overrightarrow{a}\rvert=2,\ \lvert\overrightarrow{b}\rvert=7\text{ and }\overrightarrow{a}\times\overrightarrow{b}=3\widehat{i}+2\widehat{j}+6\widehat{k},\text{ find the} \\ \text{angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta\text{ be the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.
\displaystyle \overrightarrow{a}\times\overrightarrow{b}=3\widehat{i}+2\widehat{j}+6\widehat{k}\ \text{(Given)}
\displaystyle \Rightarrow |\overrightarrow{a}\times\overrightarrow{b}|=\sqrt{3^2+2^2+6^2}
\displaystyle =\sqrt{9+4+36}
\displaystyle =7
\displaystyle \text{We know}
\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta
\displaystyle \Rightarrow 7=(2)(7)\sin\theta
\displaystyle \Rightarrow \sin\theta=\frac{1}{2}
\displaystyle \Rightarrow \theta=\frac{\pi}{6}

\displaystyle \textbf{Question 17: }~\text{What inference can you draw if }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{0}\text{ and } \\ \overrightarrow{a}\cdot\overrightarrow{b}=0?   
\displaystyle \text{Answer:}
\displaystyle \text{Given : }|\overrightarrow{a}\times\overrightarrow{b}|=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{a}=\overrightarrow{0}\ \text{or}\ \overrightarrow{b}=\overrightarrow{0}\ \text{or}\ \overrightarrow{a}\parallel\overrightarrow{b}
\displaystyle \therefore \overrightarrow{a}\parallel\overrightarrow{b}
\displaystyle \text{Also, } \overrightarrow{a}\cdot\overrightarrow{b}=0
\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\cos\theta=0
\displaystyle \Rightarrow \overrightarrow{a}=\overrightarrow{0}\ \text{or}\ \overrightarrow{b}=\overrightarrow{0}\ \text{or}\ \overrightarrow{a}\perp\overrightarrow{b}
\displaystyle \text{But a non-zero vector cannot be both parallel and perpendicular to another non-zero vector.}
\displaystyle \therefore |\overrightarrow{a}|=0\ \text{or}\ |\overrightarrow{b}|=0

\displaystyle \textbf{Question 18: }~\text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are three unit vectors such that }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{c},\\ \overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{a},\ \overrightarrow{c}\times\overrightarrow{a}=\overrightarrow{b}.\text{ Show that }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ form an orthonormal right} \\ \text{handed triad of unit vectors.}
\displaystyle \text{Answer:}
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{c}
\displaystyle \overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{a}
\displaystyle \overrightarrow{c}\times\overrightarrow{a}=\overrightarrow{b}\qquad (1)
\displaystyle \text{Now,}
\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{c}|=1\ (\because \overrightarrow{c}\text{ is a unit vector})
\displaystyle |\overrightarrow{b}\times\overrightarrow{c}|=|\overrightarrow{a}|=1\ (\because \overrightarrow{a}\text{ is a unit vector})
\displaystyle |\overrightarrow{c}\times\overrightarrow{a}|=|\overrightarrow{b}|=1\ (\because \overrightarrow{b}\text{ is a unit vector})
\displaystyle \therefore |\overrightarrow{a}\times\overrightarrow{b}|=|\overrightarrow{b}\times\overrightarrow{c}|=|\overrightarrow{c}\times\overrightarrow{a}|=1\qquad (2)
\displaystyle \text{From (1) and (2), we conclude that } \overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ form an orthonormal right handed triad of unit vectors.}

\displaystyle \textbf{Question 19: }~\text{Find a unit vector perpendicular to the plane }ABC,\text{ where the} \\ \text{coordinates of }A,B,C\text{ are }A(3,-1,2),\ B(1,-1,-3)\text{ and }C(4,-3,1).
\displaystyle \text{Answer:}
\displaystyle \text{The vector }\overrightarrow{AB}\times\overrightarrow{AC}\text{ is perpendicular to the vectors }\overrightarrow{AB}\text{ and }\overrightarrow{AC}.
\displaystyle \therefore \text{Required unit vector }=\frac{\overrightarrow{AB}\times\overrightarrow{AC}}{|\overrightarrow{AB}\times\overrightarrow{AC}|}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =( \widehat{i}-\widehat{j}-3\widehat{k})-(3\widehat{i}-\widehat{j}+2\widehat{k})
\displaystyle =-2\widehat{i}+0\widehat{j}-5\widehat{k}
\displaystyle \overrightarrow{AC}=\text{Position vector of }C-\text{Position vector of }A
\displaystyle =(4\widehat{i}-3\widehat{j}+\widehat{k})-(3\widehat{i}-\widehat{j}+2\widehat{k})
\displaystyle =\widehat{i}-2\widehat{j}-\widehat{k}
\displaystyle \therefore \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-2&0&-5\\1&-2&-1\end{vmatrix}
\displaystyle =(0\cdot(-1)-(-5)\cdot(-2))\widehat{i}-((-2)\cdot(-1)-(-5)\cdot1)\widehat{j}+((-2)\cdot(-2)-0\cdot1)\widehat{k}
\displaystyle =(0-10)\widehat{i}-(2+5)\widehat{j}+(4-0)\widehat{k}
\displaystyle =-10\widehat{i}-7\widehat{j}+4\widehat{k}
\displaystyle |\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{(-10)^2+(-7)^2+4^2}
\displaystyle =\sqrt{165}
\displaystyle \text{Unit vector perpendicular to the plane }ABC=\frac{\overrightarrow{AB}\times\overrightarrow{AC}}{|\overrightarrow{AB}\times\overrightarrow{AC}|}
\displaystyle =\frac{-10\widehat{i}-7\widehat{j}+4\widehat{k}}{\sqrt{165}}

\displaystyle \textbf{Question 20: }~\text{If }a,b,c\text{ are the lengths of sides }BC,CA\text{ and }AB\text{ of a triangle }ABC, \\ \text{ prove that }\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{0}\text{ and deduce that }\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.
\displaystyle \text{Answer:}
\displaystyle \text{We have}
\displaystyle \overrightarrow{BC}=\overrightarrow{a},\quad \overrightarrow{CA}=\overrightarrow{b},\quad \overrightarrow{AB}=\overrightarrow{c}
\displaystyle |\overrightarrow{a}|=a,\quad |\overrightarrow{b}|=b,\quad |\overrightarrow{c}|=c\ (\text{lengths are positive})
\displaystyle \text{Now,}
\displaystyle \overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{0}\ \text{(Given)}
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{a}\times(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{a}+\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{0}+\overrightarrow{a}\times\overrightarrow{b}-\overrightarrow{c}\times\overrightarrow{a}=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{c}\times\overrightarrow{a}
\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\sin C=|\overrightarrow{c}||\overrightarrow{a}|\sin B
\displaystyle \Rightarrow ab\sin C=ca\sin B
\displaystyle \Rightarrow \frac{\sin C}{c}=\frac{\sin B}{b}\qquad (1)
\displaystyle \text{Again,}
\displaystyle \overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{b}\times(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{b}\times\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{b}\times\overrightarrow{a}+\overrightarrow{b}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow -\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{0}+\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}
\displaystyle \Rightarrow |\overrightarrow{a}||\overrightarrow{b}|\sin C=|\overrightarrow{b}||\overrightarrow{c}|\sin A
\displaystyle \Rightarrow ab\sin C=bc\sin A
\displaystyle \Rightarrow \frac{\sin C}{c}=\frac{\sin A}{a}\qquad (2)
\displaystyle \text{From (1) and (2), we get}
\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}
\displaystyle \Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

\displaystyle \textbf{Question 21: }~\text{If }\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=2\widehat{i}+3\widehat{j}-5\widehat{k},\text{ then find } \\ \overrightarrow{a}\times\overrightarrow{b}.\text{ Verify that }\overrightarrow{a}\text{ and }\overrightarrow{a}\times\overrightarrow{b}\text{ are perpendicular to each other.}
\displaystyle \text{Answer:}
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b}=2\widehat{i}+3\widehat{j}-5\widehat{k}
\displaystyle \therefore \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-2&3\\2&3&-5\end{vmatrix}
\displaystyle =( (-2)\cdot(-5)-3\cdot3 )\widehat{i}-( 1\cdot(-5)-3\cdot2 )\widehat{j}+( 1\cdot3-(-2)\cdot2 )\widehat{k}
\displaystyle =(10-9)\widehat{i}-(-5-6)\widehat{j}+(3+4)\widehat{k}
\displaystyle =\widehat{i}+11\widehat{j}+7\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{a}\cdot(\overrightarrow{a}\times\overrightarrow{b})=(\widehat{i}-2\widehat{j}+3\widehat{k})\cdot(\widehat{i}+11\widehat{j}+7\widehat{k})
\displaystyle =1-22+21
\displaystyle =0
\displaystyle \text{Thus, }\overrightarrow{a}\text{ is perpendicular to }\overrightarrow{a}\times\overrightarrow{b}.

\displaystyle \textbf{Question 22: }~\text{If }\overrightarrow{p}\text{ and }\overrightarrow{q}\text{ are unit vectors forming an angle of }30^\circ, \\ \text{ find the area of the parallelogram having }\overrightarrow{a}=\overrightarrow{p}+2\overrightarrow{q}\text{ and } \\ \overrightarrow{b}=2\overrightarrow{p}+\overrightarrow{q}\text{ as its diagonals.}
\displaystyle \text{Answer:}
\displaystyle \text{Given :}
\displaystyle \overrightarrow{a}=\overrightarrow{p}+2\overrightarrow{q}
\displaystyle \overrightarrow{b}=2\overrightarrow{p}+\overrightarrow{q}
\displaystyle \overrightarrow{a}\times\overrightarrow{b}=(\overrightarrow{p}+2\overrightarrow{q})\times(2\overrightarrow{p}+\overrightarrow{q})
\displaystyle =2\overrightarrow{p}\times\overrightarrow{p}+\overrightarrow{p}\times\overrightarrow{q}+4\overrightarrow{q}\times\overrightarrow{p}+2\overrightarrow{q}\times\overrightarrow{q}
\displaystyle =2(\overrightarrow{0})+\overrightarrow{p}\times\overrightarrow{q}-4\overrightarrow{p}\times\overrightarrow{q}+2(\overrightarrow{0})
\displaystyle =-3\,\overrightarrow{p}\times\overrightarrow{q}
\displaystyle \text{Area of the parallelogram }=\frac{1}{2}\,|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =\frac{1}{2}\,|-3(\overrightarrow{p}\times\overrightarrow{q})|
\displaystyle =\frac{3}{2}\,|\overrightarrow{p}||\overrightarrow{q}|\sin30^\circ
\displaystyle =\frac{3}{2}(1)(1)\left(\frac{1}{2}\right)\ (\because\ \overrightarrow{p}\text{ and }\overrightarrow{q}\text{ are unit vectors})
\displaystyle =\frac{3}{4}\ \text{sq. units}

\displaystyle \textbf{Question 23: }~\text{For any two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ prove that } \left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2}= \begin{vmatrix} \overrightarrow{a}\cdot\overrightarrow{a} & \overrightarrow{a}\cdot\overrightarrow{b}\\ \overrightarrow{b}\cdot\overrightarrow{a} & \overrightarrow{b}\cdot\overrightarrow{b} \end{vmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=\begin{vmatrix}\overrightarrow{a}\cdot\overrightarrow{a}&\overrightarrow{a}\cdot\overrightarrow{b}\\\overrightarrow{b}\cdot\overrightarrow{a}&\overrightarrow{b}\cdot\overrightarrow{b}\end{vmatrix}
\displaystyle =\begin{vmatrix}|\overrightarrow{a}|^2&|\overrightarrow{a}||\overrightarrow{b}|\cos\theta\\|\overrightarrow{a}||\overrightarrow{b}|\cos\theta&|\overrightarrow{b}|^2\end{vmatrix}
\displaystyle =|\overrightarrow{a}|^2|\overrightarrow{b}|^2-|\overrightarrow{a}|^2|\overrightarrow{b}|^2\cos^2\theta
\displaystyle =|\overrightarrow{a}|^2|\overrightarrow{b}|^2(1-\cos^2\theta)
\displaystyle =|\overrightarrow{a}|^2|\overrightarrow{b}|^2\sin^2\theta
\displaystyle =\left(|\overrightarrow{a}||\overrightarrow{b}|\sin\theta\right)^2
\displaystyle =|\overrightarrow{a}\times\overrightarrow{b}|^2
\displaystyle =\text{LHS}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 24: }~\text{Define }\overrightarrow{a}\times\overrightarrow{b}\text{ and prove that }\left|\overrightarrow{a}\times\overrightarrow{b}\right|=(\overrightarrow{a}\cdot\overrightarrow{b})\tan\theta, \\ \text{ where }\theta\text{ is the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two non-zero, non-parallel vectors, then their vector product is defined as}
\displaystyle \overrightarrow{a}\times\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\sin\theta\,\widehat{n},
\displaystyle \text{where }\theta\text{ is the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b},
\displaystyle \text{and }\widehat{n}\text{ is the unit vector perpendicular to the plane of }\overrightarrow{a}\text{ and }\overrightarrow{b}
\displaystyle \text{such that }\overrightarrow{a},\overrightarrow{b},\widehat{n}\text{ form a right-handed system.}
\displaystyle \text{LHS}=|\overrightarrow{a}\times\overrightarrow{b}|
\displaystyle =|\overrightarrow{a}||\overrightarrow{b}|\sin\theta
\displaystyle =|\overrightarrow{a}||\overrightarrow{b}|\sin\theta\times\frac{\cos\theta}{\cos\theta}
\displaystyle =|\overrightarrow{a}||\overrightarrow{b}|\cos\theta\,\tan\theta
\displaystyle =(\overrightarrow{a}\cdot\overrightarrow{b})\tan\theta
\displaystyle =\text{RHS}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 25: }~\text{If }\lvert\overrightarrow{a}\rvert=\sqrt{26},\ \lvert\overrightarrow{b}\rvert=7\text{ and }\lvert\overrightarrow{a}\times\overrightarrow{b}\rvert=35,\text{ find }\overrightarrow{a}\cdot\overrightarrow{b}. [\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle \text{We know}
\displaystyle (\overrightarrow{a}\cdot\overrightarrow{b})^{2}+|\overrightarrow{a}\times\overrightarrow{b}|^{2}=|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}
\displaystyle \Rightarrow (\overrightarrow{a}\cdot\overrightarrow{b})^{2}+35^{2}=(\sqrt{26})^{2}\times 7^{2}\;(\because |\overrightarrow{a}\times\overrightarrow{b}|=35,\;|\overrightarrow{a}|=\sqrt{26},\;|\overrightarrow{b}|=7)
\displaystyle \Rightarrow (\overrightarrow{a}\cdot\overrightarrow{b})^{2}+1225=1274
\displaystyle \Rightarrow (\overrightarrow{a}\cdot\overrightarrow{b})^{2}=49
\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=7

\displaystyle \textbf{Question 26: }~\text{Find the area of the triangle formed by }O,A,B\text{ when } \\ \overrightarrow{OA}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{OB}=-3\widehat{i}-2\widehat{j}+\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \textbf{Given:}
\displaystyle \overrightarrow{OA}=\hat{i}+2\hat{j}+3\hat{k}
\displaystyle \overrightarrow{OB}=-3\hat{i}-2\hat{j}+\hat{k}
\displaystyle \overrightarrow{OA}\times\overrightarrow{OB}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & 3\\ -3 & -2 & 1 \end{vmatrix}
\displaystyle =(2-(-6))\hat{i}-(1-(-9))\hat{j}+(-2+6)\hat{k}
\displaystyle =8\hat{i}-10\hat{j}+4\hat{k}
\displaystyle |\overrightarrow{OA}\times\overrightarrow{OB}| =\sqrt{8^{2}+(-10)^{2}+4^{2}}
\displaystyle =\sqrt{64+100+16}
\displaystyle =\sqrt{180}=6\sqrt{5}
\displaystyle \text{Area of the triangle} =\frac{1}{2}\,|\overrightarrow{OA}\times\overrightarrow{OB}|
\displaystyle =\frac{1}{2}(6\sqrt{5})
\displaystyle =3\sqrt{5}\ \text{sq. units}

\displaystyle \textbf{Question 27: }~\text{Let }\overrightarrow{a}=\widehat{i}+4\widehat{j}+2\widehat{k},\ \overrightarrow{b}=3\widehat{i}-2\widehat{j}+7\widehat{k}\text{ and }\overrightarrow{c}=2\widehat{i}-\widehat{j}+4\widehat{k}.\text{ Find a vector }\overrightarrow{d}\text{ which is perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ and }\overrightarrow{c}\cdot\overrightarrow{d}=15. \ [\text{CBSE 2010, 2012, 2015}]
\displaystyle \text{Answer:}
\displaystyle \textbf{Given:}
\displaystyle \overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k}
\displaystyle \overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k}
\displaystyle \overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k}
\displaystyle \text{Since }\overrightarrow{d}\text{ is perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b},\text{ it is parallel to }\overrightarrow{a}\times\overrightarrow{b}.
\displaystyle \text{Suppose }\overrightarrow{d}=\lambda(\overrightarrow{a}\times\overrightarrow{b})\text{ for some scalar }\lambda.
\displaystyle \overrightarrow{a}\times\overrightarrow{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 4 & 2\\ 3 & -2 & 7 \end{vmatrix}
\displaystyle =(28+4)\hat{i}-(7-6)\hat{j}+(-2-12)\hat{k}
\displaystyle =32\hat{i}-\hat{j}-14\hat{k}
\displaystyle \overrightarrow{d}=\lambda(32\hat{i}-\hat{j}-14\hat{k})
\displaystyle \overrightarrow{c}\cdot\overrightarrow{d}=15\ (\text{Given})
\displaystyle (2\hat{i}-\hat{j}+4\hat{k})\cdot\lambda(32\hat{i}-\hat{j}-14\hat{k})=15
\displaystyle \lambda(64+1-56)=15
\displaystyle \lambda\cdot 9=15
\displaystyle \lambda=\frac{5}{3}
\displaystyle \therefore\ \overrightarrow{d}=\frac{5}{3}(32\hat{i}-\hat{j}-14\hat{k})
\displaystyle \overrightarrow{d}=\frac{1}{3}(160\hat{i}-5\hat{j}-70\hat{k})

\displaystyle \textbf{Question 28: }~\text{Find a unit vector perpendicular to each of the vectors }\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-\overrightarrow{b},\ \text{where }\overrightarrow{a}=3\widehat{i}+2\widehat{j}+2\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k}. [\text{CBSE 2011}]
\displaystyle \text{Answer:}
\displaystyle \textbf{Given:}
\displaystyle \overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}
\displaystyle \overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}
\displaystyle \overrightarrow{a}+\overrightarrow{b}=4\hat{i}+4\hat{j}+0\hat{k}
\displaystyle \overrightarrow{a}-\overrightarrow{b}=2\hat{i}+0\hat{j}+4\hat{k}
\displaystyle (\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{a}-\overrightarrow{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 4 & 4 & 0\\ 2 & 0 & 4 \end{vmatrix}
\displaystyle =\hat{i}(16-0)-\hat{j}(16-0)+\hat{k}(0-8)
\displaystyle =16\hat{i}-16\hat{j}-8\hat{k}
\displaystyle |(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{a}-\overrightarrow{b})| =\sqrt{16^{2}+(-16)^{2}+(-8)^{2}}
\displaystyle =\sqrt{256+256+64}
\displaystyle =\sqrt{576}=24
\displaystyle \text{Unit vector perpendicular to both }\overrightarrow{a}+\overrightarrow{b} \text{ and }\overrightarrow{a}-\overrightarrow{b}
\displaystyle =\frac{(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{a}-\overrightarrow{b})} {|(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{a}-\overrightarrow{b})|}
\displaystyle =\frac{16\hat{i}-16\hat{j}-8\hat{k}}{24}
\displaystyle =\frac{8(2\hat{i}-2\hat{j}-\hat{k})}{24}
\displaystyle =\frac{1}{3}(2\hat{i}-2\hat{j}-\hat{k})

\displaystyle \textbf{Question 29: }~\text{Using vectors find the area of the triangle with vertices }A(2,3,5),\\ B(3,5,8)\text{ and }C(2,7,8). [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \textbf{Let }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ be the position vectors of }A,B,C\text{ respectively.}
\displaystyle \overrightarrow{a}=2\hat{i}+3\hat{j}+5\hat{k}
\displaystyle \overrightarrow{b}=3\hat{i}+5\hat{j}+8\hat{k}
\displaystyle \overrightarrow{c}=2\hat{i}+7\hat{j}+8\hat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}
\displaystyle =(3-2)\hat{i}+(5-3)\hat{j}+(8-5)\hat{k}
\displaystyle =\hat{i}+2\hat{j}+3\hat{k}
\displaystyle \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a}
\displaystyle =(2-2)\hat{i}+(7-3)\hat{j}+(8-5)\hat{k}
\displaystyle =0\hat{i}+4\hat{j}+3\hat{k}
\displaystyle \overrightarrow{AB}\times\overrightarrow{AC}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & 3\\ 0 & 4 & 3 \end{vmatrix}
\displaystyle =\hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0)
\displaystyle =-6\hat{i}-3\hat{j}+4\hat{k}
\displaystyle |\overrightarrow{AB}\times\overrightarrow{AC}| =\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}
\displaystyle =\sqrt{36+9+16}
\displaystyle =\sqrt{61}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\,|\overrightarrow{AB}\times\overrightarrow{AC}|
\displaystyle =\frac{\sqrt{61}}{2}\ \text{sq. units}

\displaystyle \textbf{Question 30: }~\text{If }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+\widehat{k},\ \overrightarrow{b}=-\widehat{i}+\widehat{k},\ \overrightarrow{c}=2\widehat{j}-\widehat{k}\text{ are three vectors, find the} \\ \text{area of the parallelogram having diagonals }(\overrightarrow{a}+\overrightarrow{b})\text{ and }(\overrightarrow{b}+\overrightarrow{c}). [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{It is given that } \overrightarrow{a}=2\widehat{i}-3\widehat{j}+\widehat{k},\ \overrightarrow{b}=-\widehat{i}+\widehat{k},\ \overrightarrow{c}=2\widehat{j}-\widehat{k}.
\displaystyle \therefore\ \overrightarrow{a}+\overrightarrow{b}=(2\widehat{i}-3\widehat{j}+\widehat{k})+(-\widehat{i}+\widehat{k})=\widehat{i}-3\widehat{j}+2\widehat{k}.
\displaystyle \overrightarrow{b}+\overrightarrow{c}=(-\widehat{i}+\widehat{k})+(2\widehat{j}-\widehat{k})=-\widehat{i}+2\widehat{j}.
\displaystyle \text{We know that the area of a parallelogram is } \frac{1}{2}\left|\overrightarrow{d_1}\times\overrightarrow{d_2}\right|,\ \text{where } \overrightarrow{d_1} \text{ and } \overrightarrow{d_2} \text{ are the diagonal vectors.}
\displaystyle \text{Now, } (\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{b}+\overrightarrow{c})=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix}.
\displaystyle =\widehat{i}((-3)(0)-(2)(2))-\widehat{j}((1)(0)-(2)(-1))+\widehat{k}((1)(2)-(-3)(-1)).
\displaystyle =-4\widehat{i}-2\widehat{j}-\widehat{k}.
\displaystyle \text{Area of the parallelogram having diagonals } (\overrightarrow{a}+\overrightarrow{b}) \text{ and } (\overrightarrow{b}+\overrightarrow{c})
\displaystyle =\frac{1}{2}\left|(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{b}+\overrightarrow{c})\right|.
\displaystyle =\frac{1}{2}\left|-4\widehat{i}-2\widehat{j}-\widehat{k}\right|.
\displaystyle =\frac{1}{2}\sqrt{(-4)^2+(-2)^2+(-1)^2}.
\displaystyle =\frac{\sqrt{21}}{2}\ \text{square units}.
\displaystyle \text{Thus, the required area of the parallelogram is } \frac{\sqrt{21}}{2}\ \text{square units}.

\displaystyle \textbf{Question 31: }~\text{The two adjacent sides of a parallelogram are }2\widehat{i}-4\widehat{j}+5\widehat{k} \\ \text{ and }\widehat{i}-2\widehat{j}-3\widehat{k}.\text{ Find the unit vector parallel to one of its diagonals.} \\ \text{Also, find its area. }[\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle \text{Suppose }ABCD\text{ is the given parallelogram and }AC\text{ is its diagonal.}
\displaystyle \text{Let } \overrightarrow{AB}=2\widehat{i}-4\widehat{j}+5\widehat{k}.
\displaystyle \overrightarrow{BC}=\widehat{i}-2\widehat{j}-3\widehat{k}.
\displaystyle \therefore\ \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}.
\displaystyle =(2\widehat{i}-4\widehat{j}+5\widehat{k})+(\widehat{i}-2\widehat{j}-3\widehat{k}).
\displaystyle =3\widehat{i}-6\widehat{j}+2\widehat{k}.
\displaystyle |\overrightarrow{AC}|=\sqrt{3^2+(-6)^2+2^2}.
\displaystyle =\sqrt{9+36+4}.
\displaystyle =7.
\displaystyle \text{Unit vector parallel to } \overrightarrow{AC}=\frac{\overrightarrow{AC}}{|\overrightarrow{AC}|}.
\displaystyle =\frac{3\widehat{i}-6\widehat{j}+2\widehat{k}}{7}.
\displaystyle \text{Now, } \overrightarrow{AB}\times\overrightarrow{BC}=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix}.
\displaystyle =\widehat{i}((-4)(-3)-(5)(-2))-\widehat{j}((2)(-3)-(5)(1))+\widehat{k}((2)(-2)-(-4)(1)).
\displaystyle =22\widehat{i}+11\widehat{j}+0\widehat{k}.
\displaystyle |\overrightarrow{AB}\times\overrightarrow{BC}|=\sqrt{22^2+11^2+0^2}.
\displaystyle =\sqrt{484+121}.
\displaystyle =\sqrt{605}.
\displaystyle =11\sqrt{5}.
\displaystyle \text{Area of } \triangle ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{BC}|.
\displaystyle =\frac{11\sqrt{5}}{2}\ \text{sq. units}.

\displaystyle \textbf{Question 32: }~\text{If either }\overrightarrow{a}=\overrightarrow{0}\text{ or }\overrightarrow{b}=\overrightarrow{0},\text{ then }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{0}.\text{ Is the converse true?} \\ \text{Justify your answer with an example. }
\displaystyle \text{Answer:}
\displaystyle \text{If } \overrightarrow{a}=\overrightarrow{0} \text{ or } \overrightarrow{b}=\overrightarrow{0}, \text{ then } |\overrightarrow{a}||\overrightarrow{b}|\sin\theta=0.
\displaystyle \Rightarrow\ \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{0}.
\displaystyle \text{But the converse is not true, as whenever } \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{0}, \text{ we cannot be sure that either } \overrightarrow{a}=\overrightarrow{0} \text{ or } \overrightarrow{b}=\overrightarrow{0}.
\displaystyle \text{For example:}
\displaystyle \overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k}.
\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}.
\displaystyle \text{Here, } \overrightarrow{a}\neq\overrightarrow{0}.
\displaystyle \overrightarrow{b}\neq\overrightarrow{0}.
\displaystyle \text{But } \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{vmatrix}.
\displaystyle =\widehat{i}((2)(3)-(3)(2))-\widehat{j}((1)(3)-(3)(1))+\widehat{k}((1)(2)-(2)(1)).
\displaystyle =0\widehat{i}+0\widehat{j}+0\widehat{k}.
\displaystyle =\overrightarrow{0}.

\displaystyle \textbf{Question 33: }~\text{If }\overrightarrow{a}=a_{1}\widehat{i}+a_{2}\widehat{j}+a_{3}\widehat{k},\ \overrightarrow{b}=b_{1}\widehat{i}+b_{2}\widehat{j}+b_{3}\widehat{k}\text{ and } \\ \overrightarrow{c}=c_{1}\widehat{i}+c_{2}\widehat{j}+c_{3}\widehat{k},\text{ then verify that} \overrightarrow{a}\times(\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{c}. 
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle \overrightarrow{a}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}.
\displaystyle \overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}.
\displaystyle \overrightarrow{c}=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k}.
\displaystyle \overrightarrow{b}+\overrightarrow{c}=(b_1+c_1)\widehat{i}+(b_2+c_2)\widehat{j}+(b_3+c_3)\widehat{k}.
\displaystyle \therefore\ \overrightarrow{a}\times(\overrightarrow{b}+\overrightarrow{c})=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3 \end{vmatrix}.
\displaystyle =(a_2b_3+a_2c_3-a_3b_2-a_3c_2)\widehat{i}-(a_1b_3+a_1c_3-a_3b_1-a_3c_1)\widehat{j}+(a_1b_2+a_1c_2-a_2b_1-a_2c_1)\widehat{k}.\qquad(1)
\displaystyle \text{Now, } \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}.
\displaystyle =(a_2b_3-a_3b_2)\widehat{i}-(a_1b_3-a_3b_1)\widehat{j}+(a_1b_2-a_2b_1)\widehat{k}.
\displaystyle \overrightarrow{a}\times\overrightarrow{c}=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix}.
\displaystyle =(a_2c_3-a_3c_2)\widehat{i}-(a_1c_3-a_3c_1)\widehat{j}+(a_1c_2-a_2c_1)\widehat{k}.
\displaystyle \overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{c}=(a_2b_3+a_2c_3-a_3b_2-a_3c_2)\widehat{i}-(a_1b_3+a_1c_3-a_3b_1-a_3c_1)\widehat{j}+(a_1b_2+a_1c_2-a_2b_1-a_2c_1)\widehat{k}.\qquad(2)
\displaystyle \text{From (1) and (2), we get } \overrightarrow{a}\times(\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{c}.

\displaystyle \textbf{Question 34: }~\text{Using vectors, find the area of the triangle with vertices:}\\ \text{(i) }A(1,1,2),\ B(2,3,5)\text{ and }C(1,5,5)\\ \text{(ii) }A(1,2,3),\ B(2,-1,4)\text{ and }C(4,5,-1). 
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{The vertices of the triangle are } A(1,1,2),\ B(2,3,5) \text{ and } C(1,5,5).
\displaystyle \text{Position vector of } A=\widehat{i}+\widehat{j}+2\widehat{k}.
\displaystyle \text{Position vector of } B=2\widehat{i}+3\widehat{j}+5\widehat{k}.
\displaystyle \overrightarrow{AB}=(2\widehat{i}+3\widehat{j}+5\widehat{k})-(\widehat{i}+\widehat{j}+2\widehat{k}).
\displaystyle =\widehat{i}+2\widehat{j}+3\widehat{k}.
\displaystyle \text{Position vector of } C=\widehat{i}+5\widehat{j}+5\widehat{k}.
\displaystyle \overrightarrow{AC}=(\widehat{i}+5\widehat{j}+5\widehat{k})-(\widehat{i}+\widehat{j}+2\widehat{k}).
\displaystyle =4\widehat{j}+3\widehat{k}.
\displaystyle \text{Now, } \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix}.
\displaystyle =\widehat{i}((2)(3)-(3)(4))-\widehat{j}((1)(3)-(3)(0))+\widehat{k}((1)(4)-(2)(0)).
\displaystyle =-6\widehat{i}-3\widehat{j}+4\widehat{k}.
\displaystyle \text{Area of } \triangle ABC=\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|.
\displaystyle =\frac{1}{2}\left|-6\widehat{i}-3\widehat{j}+4\widehat{k}\right|.
\displaystyle =\frac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}.
\displaystyle =\frac{1}{2}\sqrt{36+9+16}.
\displaystyle =\frac{\sqrt{61}}{2}\ \text{square units}.
\displaystyle \text{(ii) }
\displaystyle \text{The vertices of the triangle are } A(1,2,3),\ B(2,-1,4) \text{ and } C(4,5,-1).
\displaystyle \text{Position vector of } A=\widehat{i}+2\widehat{j}+3\widehat{k}.
\displaystyle \text{Position vector of } B=2\widehat{i}-\widehat{j}+4\widehat{k}.
\displaystyle \text{Position vector of } C=4\widehat{i}+5\widehat{j}-\widehat{k}.
\displaystyle \overrightarrow{AB}=(2\widehat{i}-\widehat{j}+4\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k}).
\displaystyle =\widehat{i}-3\widehat{j}+\widehat{k}.
\displaystyle \overrightarrow{AC}=(4\widehat{i}+5\widehat{j}-\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k}).
\displaystyle =3\widehat{i}+3\widehat{j}-4\widehat{k}.
\displaystyle \text{Now, } \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}.
\displaystyle =\widehat{i}((-3)(-4)-(1)(3))-\widehat{j}((1)(-4)-(1)(3))+\widehat{k}((1)(3)-(-3)(3)).
\displaystyle =9\widehat{i}+7\widehat{j}+12\widehat{k}.
\displaystyle \text{Area of } \triangle ABC=\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|.
\displaystyle =\frac{1}{2}\left|9\widehat{i}+7\widehat{j}+12\widehat{k}\right|.
\displaystyle =\frac{1}{2}\sqrt{9^2+7^2+12^2}.
\displaystyle =\frac{1}{2}\sqrt{81+49+144}.
\displaystyle =\frac{\sqrt{274}}{2}\ \text{square units}.

\displaystyle \textbf{Question 35: }~\text{Find all vectors of magnitude }10\sqrt{3}\text{ that are perpendicular} \\ \text{to the plane of }\widehat{i}+2\widehat{j}+\widehat{k}\text{ and }-\widehat{i}+3\widehat{j}+4\widehat{k}. 
\displaystyle \text{Answer:}
\displaystyle \text{Let } \overrightarrow{a}=\widehat{i}+2\widehat{j}+\widehat{k} \text{ and } \overrightarrow{b}=-\widehat{i}+3\widehat{j}+4\widehat{k}.
\displaystyle \text{Unit vectors perpendicular to both } \overrightarrow{a} \text{ and } \overrightarrow{b}=\pm\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}.
\displaystyle \text{Now, } \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & 1 \\ -1 & 3 & 4 \end{vmatrix}.
\displaystyle =\widehat{i}((2)(4)-(1)(3))-\widehat{j}((1)(4)-(1)(-1))+\widehat{k}((1)(3)-(2)(-1)).
\displaystyle =5\widehat{i}-5\widehat{j}+5\widehat{k}.
\displaystyle \therefore\ |\overrightarrow{a}\times\overrightarrow{b}|=\left|5\widehat{i}-5\widehat{j}+5\widehat{k}\right|=\sqrt{5^2+(-5)^2+5^2}=\sqrt{75}=5\sqrt{3}.
\displaystyle \text{Unit vectors perpendicular to both } \overrightarrow{a} \text{ and } \overrightarrow{b}=\pm\frac{5\widehat{i}-5\widehat{j}+5\widehat{k}}{5\sqrt{3}}=\pm\frac{\widehat{i}-\widehat{j}+\widehat{k}}{\sqrt{3}}.
\displaystyle \therefore\ \text{Required vectors}=10\sqrt{3}\left(\pm\frac{\widehat{i}-\widehat{j}+\widehat{k}}{\sqrt{3}}\right)=\pm 10(\widehat{i}-\widehat{j}+\widehat{k}).
\displaystyle \text{Thus, the vectors of magnitude }10\sqrt{3}\text{ that are perpendicular to the plane of } \widehat{i}+2\widehat{j}+\widehat{k} \text{ and } -\widehat{i}+3\widehat{j}+4\widehat{k} \text{ are } \pm 10(\widehat{i}-\widehat{j}+\widehat{k}).

\displaystyle \textbf{Question 36: }~\text{The two adjacent sides of a parallelogram are }2\widehat{i}-4\widehat{j}-5\widehat{k}\text{ and } \\ 2\widehat{i}+2\widehat{j}+3\widehat{k}.\text{ Find the two unit vectors parallel to its diagonals. Using the diagonal} \\ \text{vectors,find the area of the parallelogram.} 
\displaystyle \text{Answer:}
\displaystyle \text{The two adjacent sides of a parallelogram are } 2\widehat{i}-4\widehat{j}-5\widehat{k} \text{ and } 2\widehat{i}+2\widehat{j}+3\widehat{k}.
\displaystyle \text{Suppose } \overrightarrow{a}=2\widehat{i}-4\widehat{j}-5\widehat{k} \text{ and } \overrightarrow{b}=2\widehat{i}+2\widehat{j}+3\widehat{k}.
\displaystyle \text{Then any one diagonal of the parallelogram is } \overrightarrow{p}=\overrightarrow{a}+\overrightarrow{b}.
\displaystyle \overrightarrow{p}=(2\widehat{i}-4\widehat{j}-5\widehat{k})+(2\widehat{i}+2\widehat{j}+3\widehat{k}).
\displaystyle =4\widehat{i}-2\widehat{j}-2\widehat{k}.
\displaystyle \text{Therefore, unit vector along this diagonal is } \frac{\overrightarrow{p}}{|\overrightarrow{p}|}.
\displaystyle =\frac{4\widehat{i}-2\widehat{j}-2\widehat{k}}{\sqrt{4^2+(-2)^2+(-2)^2}}.
\displaystyle =\frac{4\widehat{i}-2\widehat{j}-2\widehat{k}}{\sqrt{16+4+4}}.
\displaystyle =\frac{2\widehat{i}-\widehat{j}-\widehat{k}}{\sqrt{6}}.
\displaystyle \text{Another diagonal of the parallelogram is } \overrightarrow{p}=\overrightarrow{b}-\overrightarrow{a}.
\displaystyle \overrightarrow{p}=(2\widehat{i}+2\widehat{j}+3\widehat{k})-(2\widehat{i}-4\widehat{j}-5\widehat{k}).
\displaystyle =6\widehat{j}+8\widehat{k}.
\displaystyle \text{Therefore, unit vector along this diagonal is } \frac{\overrightarrow{p}}{|\overrightarrow{p}|}.
\displaystyle =\frac{6\widehat{j}+8\widehat{k}}{\sqrt{6^2+8^2}}.
\displaystyle =\frac{6\widehat{j}+8\widehat{k}}{\sqrt{36+64}}.
\displaystyle =\frac{3\widehat{j}+4\widehat{k}}{5}.
\displaystyle \text{Now, } (\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{b}-\overrightarrow{a})=\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 4 & -2 & -2 \\ 0 & 6 & 8 \end{vmatrix}.
\displaystyle =\widehat{i}((-2)(8)-(-2)(6))-\widehat{j}((4)(8)-(-2)(0))+\widehat{k}((4)(6)-(-2)(0)).
\displaystyle =-4\widehat{i}-32\widehat{j}+24\widehat{k}.
\displaystyle \text{Area of the parallelogram}=\frac{1}{2}\left|(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{b}-\overrightarrow{a})\right|.
\displaystyle =\frac{1}{2}\sqrt{(-4)^2+(-32)^2+24^2}.
\displaystyle =\frac{1}{2}\sqrt{16+1024+576}.
\displaystyle =\frac{\sqrt{1616}}{2}.
\displaystyle =2\sqrt{101}.

\displaystyle \textbf{Question 37: }~\text{If }\left|\overrightarrow{a}\times\overrightarrow{b}\right|^{2}+\left|\overrightarrow{a}\cdot\overrightarrow{b}\right|^{2}=400\text{ and }\lvert\overrightarrow{a}\rvert=5,\text{ then write the} \\ \text{value of }\lvert\overrightarrow{b}\rvert. \ [\text{CBSE 2016}]
\displaystyle \text{Answer:}
\displaystyle |\overrightarrow{a}\times\overrightarrow{b}|^2+|\overrightarrow{a}\cdot\overrightarrow{b}|^2=400.
\displaystyle \Rightarrow \left(|\overrightarrow{a}||\overrightarrow{b}|\sin\theta\right)^2+\left(|\overrightarrow{a}||\overrightarrow{b}|\cos\theta\right)^2=400.
\displaystyle \Rightarrow |\overrightarrow{a}|^2|\overrightarrow{b}|^2\sin^2\theta+|\overrightarrow{a}|^2|\overrightarrow{b}|^2\cos^2\theta=400.
\displaystyle \Rightarrow |\overrightarrow{a}|^2|\overrightarrow{b}|^2(\sin^2\theta+\cos^2\theta)=400.
\displaystyle \Rightarrow |\overrightarrow{a}|^2|\overrightarrow{b}|^2=400.
\displaystyle \Rightarrow 25\,|\overrightarrow{b}|^2=400.
\displaystyle \Rightarrow |\overrightarrow{b}|^2=16.
\displaystyle \Rightarrow |\overrightarrow{b}|=4.

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