Class 12: Scalar Triple Product

$\displaystyle \textbf{Question 1: Evaluate the following:}$
$\displaystyle \text{(i) }\left[\widehat{i}\ \widehat{j}\ \widehat{k}\right]+\left[\widehat{j}\ \widehat{k}\ \widehat{i}\right]+\left[\widehat{k}\ \widehat{i}\ \widehat{j}\right]$
$\displaystyle \text{Answer: }$
$\displaystyle \text{We have}$
$\displaystyle [\widehat{i}\,\widehat{j}\,\widehat{k}] + [\widehat{j}\,\widehat{k}\,\widehat{i}] + [\widehat{k}\,\widehat{i}\,\widehat{j}]$
$\displaystyle = [\widehat{i}\,\widehat{j}\,\widehat{k}] + [\widehat{i}\,\widehat{j}\,\widehat{k}] + [\widehat{i}\,\widehat{j}\,\widehat{k}] \ (\text{since } [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}] = [\overrightarrow{b}\,\overrightarrow{c}\,\overrightarrow{a}] = [\overrightarrow{c}\,\overrightarrow{a}\,\overrightarrow{b}])$
$\displaystyle = 3[\widehat{i}\,\widehat{j}\,\widehat{k}]$
$\displaystyle = 3\big((\widehat{i}\times\widehat{j})\cdot\widehat{k}\big)\ (\text{since } [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}] = (\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c})$
$\displaystyle = 3(\widehat{k}\cdot\widehat{k})$
$\displaystyle = 3(1) = 3$
$\displaystyle \text{(ii) }\left[2\widehat{i}\ \widehat{j}\ \widehat{k}\right]+\left[\widehat{i}\ \widehat{k}\ \widehat{j}\right]+\left[\widehat{k}\ \widehat{j}\ 2\widehat{i}\right]$
$\displaystyle \text{Answer: }$
$\displaystyle \text{We have}$
$\displaystyle [2\widehat{i}\,\widehat{j}\,\widehat{k}] + [\widehat{i}\,\widehat{k}\,\widehat{j}] + [\widehat{k}\,\widehat{j}\,2\widehat{i}]$
$\displaystyle = 2[\widehat{i}\,\widehat{j}\,\widehat{k}] + [\widehat{i}\,\widehat{k}\,\widehat{j}] + 2[\widehat{k}\,\widehat{j}\,\widehat{i}] \ (\text{since } [l\overrightarrow{a}\,m\overrightarrow{b}\,n\overrightarrow{c}] = lmn[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}])$
$\displaystyle = 2\big((\widehat{i}\times\widehat{j})\cdot\widehat{k}\big) + \big((\widehat{i}\times\widehat{k})\cdot\widehat{j}\big) + 2\big((\widehat{k}\times\widehat{j})\cdot\widehat{i}\big)\ (\text{since } [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}] = (\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c})$
$\displaystyle = 2(\widehat{k}\cdot\widehat{k}) + ((-\widehat{j})\cdot\widehat{j}) + 2((-\widehat{i})\cdot\widehat{i})$
$\displaystyle = 2(1) + (-1) + 2(-1)$
$\displaystyle = 2 - 1 - 2 = -1$

$\displaystyle \textbf{Question 2: Find }[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}],\ \text{when}$
$\displaystyle \text{(i) }\overrightarrow{a}=2\widehat{i}-3\widehat{j},\ \overrightarrow{b}=\widehat{i}+\widehat{j}-\widehat{k},\ \overrightarrow{c}=3\widehat{i}-\widehat{k}$
$\displaystyle \text{Answer: }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=2\widehat{i}-3\widehat{j}$
$\displaystyle \overrightarrow{b}=\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \overrightarrow{c}=3\widehat{i}-\widehat{k}$
$\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=(2\widehat{i}-3\widehat{j})\times(\widehat{i}+\widehat{j}-\widehat{k})$
$\displaystyle =2\widehat{i}\times\widehat{i}+2\widehat{i}\times\widehat{j}-2\widehat{i}\times\widehat{k}-3\widehat{j}\times\widehat{i}-3\widehat{j}\times\widehat{j}+3\widehat{j}\times\widehat{k}$
$\displaystyle =0+2\widehat{k}+2\widehat{j}+3\widehat{k}+3\widehat{i}$
$\displaystyle =3\widehat{i}+2\widehat{j}+5\widehat{k}$
$\displaystyle (\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c}=(3\widehat{i}+2\widehat{j}+5\widehat{k})\cdot(3\widehat{i}-\widehat{k})$
$\displaystyle =9-5=4\ \ldots (1)$
$\displaystyle \text{Now,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=(\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c}$
$\displaystyle =4\ [\text{Using }(1)]$
$\displaystyle \text{(ii) } \overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=2\widehat{i}+\widehat{j}-\widehat{k},\ \overrightarrow{c}=\widehat{j}+\widehat{k}$
$\displaystyle \text{Answer: }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{b}=2\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \overrightarrow{c}=\widehat{j}+\widehat{k}$
$\displaystyle \therefore \ \overrightarrow{a}\times\overrightarrow{b}=(\widehat{i}-2\widehat{j}+3\widehat{k})\times(2\widehat{i}+\widehat{j}-\widehat{k})$
$\displaystyle =\widehat{i}\times2\widehat{i}+\widehat{i}\times\widehat{j}-\widehat{i}\times\widehat{k}-2\widehat{j}\times2\widehat{i}-2\widehat{j}\times\widehat{j}+2\widehat{j}\times\widehat{k}+3\widehat{k}\times2\widehat{i}+3\widehat{k}\times\widehat{j}-3\widehat{k}\times\widehat{k}$
$\displaystyle =0+\widehat{k}+\widehat{j}+4\widehat{k}+0+2\widehat{i}+6\widehat{j}-3\widehat{i}+0$
$\displaystyle =-\widehat{i}+7\widehat{j}+5\widehat{k}$
$\displaystyle (\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c}=(-\widehat{i}+7\widehat{j}+5\widehat{k})\cdot(\widehat{j}+\widehat{k})$
$\displaystyle =7+5=12\ \ldots (1)$
$\displaystyle \text{Now,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=(\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c}$
$\displaystyle =12\ [\text{Using }(1)]$

$\displaystyle \textbf{Question 3: Find the volume of the parallelepiped whose coterminous} \\ \text{edges are represented by the vectors:}$
$\displaystyle \text{(i) }\overrightarrow{a}=2\widehat{i}+3\widehat{j}+4\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{c}=3\widehat{i}-\widehat{j}+2\widehat{k}$
$\displaystyle \text{(ii) }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+4\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{c}=3\widehat{i}-\widehat{j}-2\widehat{k}$
$\displaystyle \text{(iii) }\overrightarrow{a}=11\widehat{i},\ \overrightarrow{b}=2\widehat{j},\ \overrightarrow{c}=13\widehat{k}$
$\displaystyle \text{(iv) }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{c}=\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=2\widehat{i}+3\widehat{j}+4\widehat{k}$
$\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \overrightarrow{c}=3\widehat{i}-\widehat{j}+2\widehat{k}$
$\displaystyle \text{We know that the volume of a parallelepiped whose three adjacent edges are }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \\ \text{ is equal to }|[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]|.$
$\displaystyle \text{Here,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=\begin{vmatrix}2&3&4\\1&2&-1\\3&-1&2\end{vmatrix}$
$\displaystyle =2(2\cdot2-(-1)\cdot(-1))-3(1\cdot2-(-1)\cdot3)+4(1\cdot(-1)-2\cdot3)$
$\displaystyle =2(4-1)-3(2-(-3))+4(-1-6)$
$\displaystyle =6-15-28=-37$
$\displaystyle \text{Volume of the parallelepiped }=|[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]|=|-37|=37\ \text{cubic units}$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=2\widehat{i}-3\widehat{j}+4\widehat{k}$
$\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \overrightarrow{c}=3\widehat{i}-\widehat{j}-2\widehat{k}$
$\displaystyle \text{We know that the volume of a parallelepiped whose three adjacent edges are }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \\ \text{ is equal to }|[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]|.$
$\displaystyle \text{Here,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=\begin{vmatrix}2&-3&4\\1&2&-1\\3&-1&-2\end{vmatrix}$
$\displaystyle =2(2\cdot(-2)-(-1)\cdot(-1))-(-3)(1\cdot(-2)-(-1)\cdot3)+4(1\cdot(-1)-2\cdot3)$
$\displaystyle =2(-4-1)+3(-2+3)+4(-1-6)$
$\displaystyle =-10+3-28=-35$
$\displaystyle \text{Volume of the parallelepiped }=|[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]|=|-35|=35\ \text{cubic units}$
$\displaystyle \text{(iii) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=11\widehat{i}$
$\displaystyle \overrightarrow{b}=2\widehat{j}$
$\displaystyle \overrightarrow{c}=13\widehat{k}$
$\displaystyle \text{We know that the volume of a parallelepiped whose three adjacent edges are }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \\ \text{ is equal to }|[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]|.$
$\displaystyle \text{Here,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=\begin{vmatrix}11&0&0\\0&2&0\\0&0&13\end{vmatrix}$
$\displaystyle =11(2\cdot13-0\cdot0)-0(0\cdot13-0\cdot0)+0(0\cdot0-2\cdot0)$
$\displaystyle =11(26)=286$
$\displaystyle \text{Volume of the parallelepiped }=|[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]|=|286|=286\ \text{cubic units}$
$\displaystyle \text{(iv) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{c}=\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \text{We know that the volume of a parallelepiped whose three adjacent edges are }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \\ \text{ is equal to }|[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]|.$
$\displaystyle \text{Here,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=\begin{vmatrix}1&1&1\\1&-1&1\\1&2&-1\end{vmatrix}$
$\displaystyle =1((-1)\cdot(-1)-1\cdot2)-1(1\cdot(-1)-1\cdot1)+1(1\cdot2-(-1)\cdot1)$
$\displaystyle =1(1-2)-1(-1-1)+1(2+1)$
$\displaystyle =-1+2+3=4$
$\displaystyle \text{Volume of the parallelepiped }=|[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]|=|4|=4\ \text{cubic units}$

$\displaystyle \textbf{Question 4: Show that each of the following triads of vectors are coplanar:}$
$\displaystyle \text{(i) }\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{b}=3\widehat{i}+2\widehat{j}+7\widehat{k},\ \overrightarrow{c}=5\widehat{i}+6\widehat{j}+5\widehat{k}$
$\displaystyle \text{(ii) }\overrightarrow{a}=4\widehat{i}-6\widehat{j}-2\widehat{k},\ \overrightarrow{b}=-\widehat{i}+4\widehat{j}+3\widehat{k},\ \overrightarrow{c}=-8\widehat{i}-\widehat{j}+3\widehat{k}$
$\displaystyle \text{(iii) }\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=-2\widehat{i}+3\widehat{j}-4\widehat{k},\ \overrightarrow{c}=\widehat{i}-3\widehat{j}+5\widehat{k}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \overrightarrow{b}=3\widehat{i}+2\widehat{j}+7\widehat{k}$
$\displaystyle \overrightarrow{c}=5\widehat{i}+6\widehat{j}+5\widehat{k}$
$\displaystyle \text{We know that three vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff their} \\ \text{scalar triple product is zero, i.e. }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{Here,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=\begin{vmatrix}1&2&-1\\3&2&7\\5&6&5\end{vmatrix}$
$\displaystyle =1(2\cdot5-7\cdot6)-2(3\cdot5-7\cdot5)-1(3\cdot6-2\cdot5)$
$\displaystyle =1(10-42)-2(15-35)-1(18-10)$
$\displaystyle =-32+40-8=0$
$\displaystyle \text{Hence, the given vectors are coplanar.}$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=-4\widehat{i}-6\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{b}=-\widehat{i}+4\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{c}=-8\widehat{i}-\widehat{j}+3\widehat{k}$
$\displaystyle \text{We know that three vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff their} \\ \text{scalar triple product is zero, i.e. }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{Here,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=\begin{vmatrix}-4&-6&-2\\-1&4&3\\-8&-1&3\end{vmatrix}$
$\displaystyle =-4(4\cdot3-3\cdot(-1))-(-6)((-1)\cdot3-3\cdot(-8))+(-2)((-1)\cdot(-1)-4\cdot(-8))$
$\displaystyle =-4(12+3)+6(-3+24)-2(1+32)$
$\displaystyle =-60+126-66=0$
$\displaystyle \text{Hence, the given vectors are coplanar.}$
$\displaystyle \text{(iii) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{b}=-2\widehat{i}+3\widehat{j}-4\widehat{k}$
$\displaystyle \overrightarrow{c}=\widehat{i}-3\widehat{j}+5\widehat{k}$
$\displaystyle \text{We know that three vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff their} \\ \text{scalar triple product is zero, i.e. }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{Here,}$
$\displaystyle [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=\begin{vmatrix}1&-2&3\\-2&3&-4\\1&-3&5\end{vmatrix}$
$\displaystyle =1(3\cdot5-(-4)\cdot(-3))-(-2)((-2)\cdot5-(-4)\cdot1)+3((-2)\cdot(-3)-3\cdot1)$
$\displaystyle =1(15-12)+2(-10+4)+3(6-3)$
$\displaystyle =3-12+9=0$
$\displaystyle \text{Hence, the given vectors are coplanar.}$

$\displaystyle \textbf{Question 5: Find the value of }\lambda\text{ so that the following vectors are coplanar:}$
$\displaystyle \text{(i) }\overrightarrow{a}=\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{b}=2\widehat{i}+\widehat{j}-\widehat{k},\ \overrightarrow{c}=\lambda\widehat{i}-\widehat{j}+\lambda\widehat{k}$
$\displaystyle \text{(ii) }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-3\widehat{k},\ \overrightarrow{c}=\lambda\widehat{i}+\lambda\widehat{j}+5\widehat{k}$
$\displaystyle \text{(iii) }\overrightarrow{a}=\widehat{i}+2\widehat{j}-3\widehat{k},\ \overrightarrow{b}=3\widehat{i}+\lambda\widehat{j}+\widehat{k},\ \overrightarrow{c}=\widehat{i}+2\widehat{j}+2\widehat{k}$
$\displaystyle \text{(iv) }\overrightarrow{a}=\widehat{i}+3\widehat{j},\ \overrightarrow{b}=5\widehat{k},\ \overrightarrow{c}=\lambda\widehat{i}-\widehat{j}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{(i) Given :}$
$\displaystyle \overrightarrow{a}=\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{b}=2\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \overrightarrow{c}=\lambda\widehat{i}-\widehat{j}+\lambda\widehat{k}$
$\displaystyle \text{We know that vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{It is given that }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar.}$
$\displaystyle \therefore [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0$
$\displaystyle \Rightarrow \begin{vmatrix}1&-1&1\\2&1&-1\\\lambda&-1&\lambda\end{vmatrix}=0$
$\displaystyle \Rightarrow 1(1\cdot\lambda-(-1)\cdot(-1))-(-1)(2\cdot\lambda-(-1)\cdot\lambda)+1(2\cdot(-1)-1\cdot\lambda)=0$
$\displaystyle \Rightarrow 1(\lambda-1)+1(2\lambda+\lambda)+1(-2-\lambda)=0$
$\displaystyle \Rightarrow \lambda-1+3\lambda-2-\lambda=0$
$\displaystyle \Rightarrow 3\lambda-3=0$
$\displaystyle \Rightarrow \lambda=1$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}-3\widehat{k}$
$\displaystyle \overrightarrow{c}=\lambda\widehat{i}+\lambda\widehat{j}+5\widehat{k}$
$\displaystyle \text{We know that vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{It is given that }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar.}$
$\displaystyle \therefore [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0$
$\displaystyle \Rightarrow \begin{vmatrix}2&-1&1\\1&2&-3\\\lambda&\lambda&5\end{vmatrix}=0$
$\displaystyle \Rightarrow 2(2\cdot5-(-3)\cdot\lambda)-(-1)(1\cdot5-(-3)\cdot\lambda)+1(1\cdot\lambda-2\cdot\lambda)=0$
$\displaystyle \Rightarrow 2(10+3\lambda)+1(5+3\lambda)+1(\lambda-2\lambda)=0$
$\displaystyle \Rightarrow 20+6\lambda+5+3\lambda-\lambda=0$
$\displaystyle \Rightarrow 8\lambda+25=0$
$\displaystyle \Rightarrow \lambda=-\frac{25}{8}$
$\displaystyle \text{(iii) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=\widehat{i}+2\widehat{j}-3\widehat{k}$
$\displaystyle \overrightarrow{b}=3\widehat{i}+\lambda\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{c}=\widehat{i}+2\widehat{j}+2\widehat{k}$
$\displaystyle \text{We know that vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{It is given that }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar.}$
$\displaystyle \therefore [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0$
$\displaystyle \Rightarrow \begin{vmatrix}1&2&-3\\3&\lambda&1\\1&2&2\end{vmatrix}=0$
$\displaystyle \Rightarrow 1(\lambda\cdot2-1\cdot2)-2(3\cdot2-1\cdot1)-3(3\cdot2-\lambda\cdot1)=0$
$\displaystyle \Rightarrow 1(2\lambda-2)-2(6-1)-3(6-\lambda)=0$
$\displaystyle \Rightarrow 2\lambda-2-10-18+3\lambda=0$
$\displaystyle \Rightarrow 5\lambda-30=0$
$\displaystyle \Rightarrow \lambda=6$
$\displaystyle \text{(iv) }$
$\displaystyle \text{Given :}$
$\displaystyle \overrightarrow{a}=\widehat{i}+3\widehat{j}$
$\displaystyle \overrightarrow{b}=5\widehat{k}$
$\displaystyle \overrightarrow{c}=\lambda\widehat{i}-\widehat{j}$
$\displaystyle \text{We know that vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{It is given that }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar.}$
$\displaystyle \therefore [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0$
$\displaystyle \Rightarrow \begin{vmatrix}1&3&0\\0&0&5\\\lambda&-1&0\end{vmatrix}=0$
$\displaystyle \Rightarrow 1(0\cdot0-5\cdot(-1))-3(0\cdot0-5\cdot\lambda)+0(0\cdot(-1)-0\cdot\lambda)=0$
$\displaystyle \Rightarrow 1(5)-3(-5\lambda)+0=0$
$\displaystyle \Rightarrow 5+15\lambda=0$
$\displaystyle \Rightarrow \lambda=-\frac{1}{3}$

$\displaystyle \textbf{Question 6: } \text{Show that the four points having position vectors }6\widehat{i}-7\widehat{j},\\ 16\widehat{i}-19\widehat{j}-4\widehat{k},\ 3\widehat{j}-6\widehat{k},\ 2\widehat{i}-5\widehat{j}+10\widehat{k}\text{ are not coplanar.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A,B,C\text{ and }D\text{ be the given points. These points will be coplanar iff any one} \\ \text{of the following triads of vectors are coplanar:}$
$\displaystyle \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD};\ \overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CD};\ \overrightarrow{BC},\overrightarrow{BA},\overrightarrow{BD}\ \text{etc.}$
$\displaystyle \text{To show that }\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\text{ are not coplanar, we have to} \\ \text{prove that their scalar triple product, i.e. }[\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]\neq 0.$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{PQ}=(\text{Position vector of }Q)-(\text{Position vector of }P)$
$\displaystyle \overrightarrow{AB}=(16\widehat{i}-19\widehat{j}-4\widehat{k})-(6\widehat{i}-7\widehat{j})=10\widehat{i}-12\widehat{j}-4\widehat{k}$
$\displaystyle \overrightarrow{AC}=(3\widehat{j}-6\widehat{k})-(6\widehat{i}-7\widehat{j})=-6\widehat{i}+10\widehat{j}-6\widehat{k}$
$\displaystyle \overrightarrow{AD}=(2\widehat{i}+5\widehat{j}+10\widehat{k})-(6\widehat{i}-7\widehat{j})=-4\widehat{i}+12\widehat{j}+10\widehat{k}$
$\displaystyle \therefore [\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]=\begin{vmatrix}10&-12&-4\\-6&10&-6\\-4&12&10\end{vmatrix}$
$\displaystyle =10\begin{vmatrix}10&-6\\12&10\end{vmatrix}-(-12)\begin{vmatrix}-6&-6\\-4&10\end{vmatrix}+(-4)\begin{vmatrix}-6&10\\-4&12\end{vmatrix}$
$\displaystyle =10(100+72)+12(-60-24)-4(-72+40)$
$\displaystyle =1720-1008+128=840\neq 0$
$\displaystyle \text{Thus, the given points are not coplanar.}$

$\displaystyle \textbf{Question 7: Show that the points }A(-1,4,-3),\ B(3,2,-5),\ C(-3,8,-5) \\ \text{ and }D(-3,2,1)\text{ are coplanar.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The points }A,B,C\text{ and }D\text{ will be coplanar iff any one of the following triads of} \\ \text{vectors are coplanar:}$
$\displaystyle \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD};\ \overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CD};\ \overrightarrow{BC},\overrightarrow{BA},\overrightarrow{BD}\ \text{etc.}$
$\displaystyle \text{To show that }\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\text{ are coplanar, we have to prove that their scalar triple} \\ \text{product,}$
$\displaystyle \text{i.e. }[\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]=0$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{AB}=[3-(-1)]\widehat{i}+(2-4)\widehat{j}+[-5-(-3)]\widehat{k}=4\widehat{i}-2\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{AC}=[-3-(-1)]\widehat{i}+(8-4)\widehat{j}+[-5-(-3)]\widehat{k}=-2\widehat{i}+4\widehat{j}-2\widehat{k}$
$\displaystyle \overrightarrow{AD}=[-3-(-1)]\widehat{i}+(2-4)\widehat{j}+[1-(-3)]\widehat{k}=-2\widehat{i}-2\widehat{j}+4\widehat{k}$
$\displaystyle \therefore [\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]=\begin{vmatrix}4&-2&-2\\-2&4&-2\\-2&-2&4\end{vmatrix}$
$\displaystyle =4(16-4)+2(-8-4)-2(4+8)$
$\displaystyle =4(12)+2(-12)-2(12)=48-24-24=0$
$\displaystyle \text{Thus, the given points are coplanar.}$

$\displaystyle \textbf{Question 8: Show that four points whose position vectors are }6\widehat{i}-7\widehat{j},\ 16\widehat{i}-19\widehat{j}-4\widehat{k},\ 3\widehat{i}-6\widehat{k},\ 2\widehat{i}-5\widehat{j}+10\widehat{k}\text{ are coplanar. \ [CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A,B,C\text{ and }D\text{ be the given points. The given points will be coplanar iff any} \\ \text{one of the following triads of vectors are coplanar:}$
$\displaystyle \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD};\ \overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CD};\ \overrightarrow{BC},\overrightarrow{BA},\overrightarrow{BD}\ \text{etc.}$
$\displaystyle \text{In order to show that }\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\text{ are \emph{not} coplanar, we have to show that their} \\ \text{scalar triple product}$
$\displaystyle \text{i.e. }[\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]\neq 0$
$\displaystyle \text{Using, }\overrightarrow{PQ}=(\text{Position vector of }Q)-(\text{Position vector of }P),\text{ we obtain}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{AB}=(16\widehat{i}-19\widehat{j}-4\widehat{k})-(6\widehat{i}-7\widehat{j})=10\widehat{i}-12\widehat{j}-4\widehat{k}$
$\displaystyle \overrightarrow{AC}=(3\widehat{i}-6\widehat{k})-(6\widehat{i}-7\widehat{j})=-3\widehat{i}+7\widehat{j}-6\widehat{k}$
$\displaystyle \text{and }\overrightarrow{AD}=(2\widehat{i}-5\widehat{j}+10\widehat{k})-(6\widehat{i}-7\widehat{j})=-4\widehat{i}+2\widehat{j}+10\widehat{k}$
$\displaystyle \therefore [\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]=\begin{vmatrix}10&-12&-4\\-3&7&-6\\-4&2&10\end{vmatrix}$
$\displaystyle =10(7\cdot10-(-6)\cdot2)+12((-3)\cdot10-(-6)\cdot(-4))-4((-3)\cdot2-7\cdot(-4))$
$\displaystyle =10(70+12)+12(-30-24)-4(-6+28)$
$\displaystyle =820-648-88=84\neq 0$
$\displaystyle \text{Thus, the given points are not coplanar.}$

$\displaystyle \textbf{Question 9: Find the value of }\lambda\text{ for which the four points with} \\ \text{position vectors }-\widehat{j}-\widehat{k},\ 4\widehat{i}+5\widehat{j}+\lambda\widehat{k},\ 3\widehat{i}+9\widehat{j}+4\widehat{k}\text{ and }-4\widehat{i}+4\widehat{j}+4\widehat{k}\text{ are coplanar.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A,B,C\text{ and }D\text{ be the given points. Then,}$
$\displaystyle \overrightarrow{AB}=(4\widehat{i}+5\widehat{j}+\lambda\widehat{k})-(0\widehat{i}-\widehat{j}-\widehat{k})=4\widehat{i}+6\widehat{j}+(\lambda+1)\widehat{k}$
$\displaystyle \overrightarrow{AC}=(3\widehat{i}+9\widehat{j}+4\widehat{k})-(0\widehat{i}-\widehat{j}-\widehat{k})=3\widehat{i}+10\widehat{j}+5\widehat{k}$
$\displaystyle \overrightarrow{AD}=(-4\widehat{i}+4\widehat{j}+4\widehat{k})-(0\widehat{i}-\widehat{j}-\widehat{k})=-4\widehat{i}+5\widehat{j}+5\widehat{k}$
$\displaystyle \text{The given points are coplanar iff vectors }\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\text{ are coplanar.}$
$\displaystyle \text{Now, }\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\text{ are coplanar.}$
$\displaystyle \Rightarrow [\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]=0$
$\displaystyle \Rightarrow \begin{vmatrix}4&6&\lambda+1\\3&10&5\\-4&5&5\end{vmatrix}=0$
$\displaystyle \Rightarrow 4(10\cdot5-5\cdot5)-6(3\cdot5-5\cdot(-4))+(\lambda+1)(3\cdot5-10\cdot(-4))=0$
$\displaystyle \Rightarrow 4(50-25)-6(15+20)+(\lambda+1)(15+40)=0$
$\displaystyle \Rightarrow 100-210+55\lambda+55=0$
$\displaystyle \Rightarrow 55\lambda=55$
$\displaystyle \Rightarrow \lambda=1$

$\displaystyle \textbf{Question 10: Prove that }(\overrightarrow{a}-\overrightarrow{b})\cdot\big((\overrightarrow{b}-\overrightarrow{c})\times(\overrightarrow{c}-\overrightarrow{a})\big)=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have}$
$\displaystyle (\overrightarrow{a}-\overrightarrow{b})\cdot\{(\overrightarrow{b}-\overrightarrow{c})\times(\overrightarrow{c}-\overrightarrow{a})\}$
$\displaystyle =(\overrightarrow{a}-\overrightarrow{b})\cdot(\overrightarrow{b}\times\overrightarrow{c}-\overrightarrow{b}\times\overrightarrow{a}-\overrightarrow{c}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a})\ \text{(by distributive law)}$
$\displaystyle =(\overrightarrow{a}-\overrightarrow{b})\cdot(\overrightarrow{b}\times\overrightarrow{c}-\overrightarrow{b}\times\overrightarrow{a}+\overrightarrow{c}\times\overrightarrow{a})\ \text{(since }\overrightarrow{c}\times\overrightarrow{c}=0\text{)}$
$\displaystyle =(\overrightarrow{a}-\overrightarrow{b})\cdot(\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{c}\times\overrightarrow{a})$
$\displaystyle =\overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c})+\overrightarrow{a}\cdot(\overrightarrow{a}\times\overrightarrow{b})+\overrightarrow{a}\cdot(\overrightarrow{c}\times\overrightarrow{a})-\overrightarrow{b}\cdot(\overrightarrow{b}\times\overrightarrow{c})-\overrightarrow{b}\cdot(\overrightarrow{a}\times\overrightarrow{b})-\overrightarrow{b}\cdot(\overrightarrow{c}\times\overrightarrow{a})\ \text{(by distributive law)}$
$\displaystyle =[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]+0+0-0-0-[\overrightarrow{b}\,\overrightarrow{c}\,\overrightarrow{a}]$
$\displaystyle =[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]-[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]\ \text{(since }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=[\overrightarrow{b}\,\overrightarrow{c}\,\overrightarrow{a}]\text{)}$
$\displaystyle =0$

$\displaystyle \textbf{Question 11: If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are the position vectors of points }A,B,C\text{ respectively, prove that } \\ \overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}\text{ is perpendicular to the plane of }\triangle ABC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that if any vector is perpendicular to all three sides of } \\ \triangle ABC,\text{ it must be perpendicular to the plane of }\triangle ABC.$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a},\ \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b},\ \overrightarrow{CA}=\overrightarrow{a}-\overrightarrow{c}\ (\text{position vectors of }A,B,C\text{ are }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c})$
$\displaystyle \text{We have}$
$\displaystyle \overrightarrow{AB}\cdot(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a})$
$\displaystyle =(\overrightarrow{b}-\overrightarrow{a})\cdot(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a})$
$\displaystyle =\overrightarrow{b}\cdot(\overrightarrow{a}\times\overrightarrow{b})+\overrightarrow{b}\cdot(\overrightarrow{b}\times\overrightarrow{c})+\overrightarrow{b}\cdot(\overrightarrow{c}\times\overrightarrow{a})-\overrightarrow{a}\cdot(\overrightarrow{a}\times\overrightarrow{b})-\overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c})-\overrightarrow{a}\cdot(\overrightarrow{c}\times\overrightarrow{a})\ \text{(by distributive law)}$
$\displaystyle =[\overrightarrow{b}\,\overrightarrow{a}\,\overrightarrow{b}]+[\overrightarrow{b}\,\overrightarrow{b}\,\overrightarrow{c}]+[\overrightarrow{b}\,\overrightarrow{c}\,\overrightarrow{a}]-[\overrightarrow{a}\,\overrightarrow{a}\,\overrightarrow{b}]-[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]-[\overrightarrow{a}\,\overrightarrow{c}\,\overrightarrow{a}]$
$\displaystyle =0+0+[\overrightarrow{b}\,\overrightarrow{c}\,\overrightarrow{a}]-0-[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]-0$
$\displaystyle =0\ (\because [\overrightarrow{b}\,\overrightarrow{c}\,\overrightarrow{a}]=[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}])$
$\displaystyle \overrightarrow{BC}\cdot(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a})$
$\displaystyle =(\overrightarrow{c}-\overrightarrow{b})\cdot(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a})$
$\displaystyle =\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b})+\overrightarrow{c}\cdot(\overrightarrow{b}\times\overrightarrow{c})+\overrightarrow{c}\cdot(\overrightarrow{c}\times\overrightarrow{a})-\overrightarrow{b}\cdot(\overrightarrow{a}\times\overrightarrow{b})-\overrightarrow{b}\cdot(\overrightarrow{b}\times\overrightarrow{c})-\overrightarrow{b}\cdot(\overrightarrow{c}\times\overrightarrow{a})\ \text{(by distributive law)}$
$\displaystyle =[\overrightarrow{c}\,\overrightarrow{a}\,\overrightarrow{b}]+0+0-0-0-[\overrightarrow{b}\,\overrightarrow{c}\,\overrightarrow{a}]$
$\displaystyle =0\ (\because [\overrightarrow{c}\,\overrightarrow{a}\,\overrightarrow{b}]=[\overrightarrow{b}\,\overrightarrow{c}\,\overrightarrow{a}])$
$\displaystyle \text{Similarly,}$
$\displaystyle \overrightarrow{CA}\cdot(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a})$
$\displaystyle =(\overrightarrow{a}-\overrightarrow{c})\cdot(\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a})$
$\displaystyle =\overrightarrow{a}\cdot(\overrightarrow{a}\times\overrightarrow{b})+\overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c})+\overrightarrow{a}\cdot(\overrightarrow{c}\times\overrightarrow{a})-\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b})-\overrightarrow{c}\cdot(\overrightarrow{b}\times\overrightarrow{c})-\overrightarrow{c}\cdot(\overrightarrow{c}\times\overrightarrow{a})\ \text{(by distributive law)}$
$\displaystyle =0+[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]+0-[\overrightarrow{c}\,\overrightarrow{a}\,\overrightarrow{b}]-0-0$
$\displaystyle =0\ (\because [\overrightarrow{c}\,\overrightarrow{a}\,\overrightarrow{b}]=[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}])$
$\displaystyle \text{Hence, the vector }\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}\text{ is perpendicular to all sides of } \\ \triangle ABC\text{ and also perpendicular to the plane of }\triangle ABC.$

$\displaystyle \textbf{Question 12: Let }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}\text{ and }\overrightarrow{c}=c_{1}\widehat{i}+c_{2}\widehat{j}+c_{3}\widehat{k}. \text{ Then,}$
$\displaystyle \text{(i) If }c_{1}=1\text{ and }c_{2}=2,\text{ find }c_{3}\text{ which makes }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ coplanar.}$
$\displaystyle \text{(ii) If }c_{2}=-1\text{ and }c_{3}=1,\text{ show that no value of }c_{1}\text{ can make }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ coplanar.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{If }c_1=1\text{ and }c_2=2,\text{ then }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}\text{ and }\overrightarrow{c}=\widehat{i}+2\widehat{j}+c_3\widehat{k}.$
$\displaystyle \text{We know that vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{It is given that }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar.}$
$\displaystyle \therefore [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0$
$\displaystyle \Rightarrow \begin{vmatrix}1&1&1\\1&0&0\\1&2&c_3\end{vmatrix}=0$
$\displaystyle \Rightarrow 1(0\cdot c_3-0\cdot2)-1(1\cdot c_3-0\cdot1)+1(1\cdot2-0\cdot1)=0$
$\displaystyle \Rightarrow 0-c_3+2=0$
$\displaystyle \Rightarrow c_3=2$

$\displaystyle \text{(ii) }$
$\displaystyle \text{If }c_2=-1\text{ and }c_3=1,\text{ then }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}\text{ and }\overrightarrow{c}=c_1\widehat{i}-\widehat{j}+\widehat{k}.$
$\displaystyle \text{We know that vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar iff }[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0.$
$\displaystyle \text{For }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ to be coplanar,}$
$\displaystyle \Rightarrow [\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=0$
$\displaystyle \Rightarrow \begin{vmatrix}1&1&1\\1&0&0\\c_1&-1&1\end{vmatrix}=0$
$\displaystyle \Rightarrow 1(0\cdot1-0\cdot(-1))-1(1\cdot1-0\cdot c_1)+1(1\cdot(-1)-0\cdot c_1)=0$
$\displaystyle \Rightarrow 0-1-1=0$
$\displaystyle \Rightarrow -2=0$
$\displaystyle \text{But this is not possible for any value of }c_1.$
$\displaystyle \text{Hence, no value of }c_1\text{ can make }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ coplanar.}$

$\displaystyle \textbf{Question 13: Find }\lambda\text{ for which the points }A(3,2,1),\ B(4,\lambda,5),\ C(4,2,-2) \\ \text{ and }D(6,5,-1)\text{ are coplanar. \ [CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The points }A,B,C\text{ and }D\text{ will be coplanar iff any one of the following} \\ \text{triads of vectors are coplanar:}$
$\displaystyle \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD};\ \overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CD};\ \overrightarrow{BC},\overrightarrow{BA},\overrightarrow{BD}\ \text{etc.}$
$\displaystyle \text{It is given that }\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\text{ are coplanar.}$
$\displaystyle \text{Thus, their scalar triple product }[\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]\text{ is equal to zero.}$
$\displaystyle \text{Now,}$
$\displaystyle \text{Direction ratios of }\overrightarrow{PQ}=(\text{direction ratios of }Q)-(\text{direction ratios of }P)$
$\displaystyle \text{Direction ratios of vector }\overrightarrow{AB}=(4-3,\lambda-2,5-1)\text{ i.e. }1,\lambda-2,4$
$\displaystyle \text{Direction ratios of vector }\overrightarrow{AC}=(4-3,2-2,-2-1)\text{ i.e. }1,0,-3$
$\displaystyle \text{Direction ratios of vector }\overrightarrow{AD}=(6-3,5-2,-1-1)\text{ i.e. }3,3,-2$
$\displaystyle \therefore [\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]=\begin{vmatrix}1&\lambda-2&4\\1&0&-3\\3&3&-2\end{vmatrix}$
$\displaystyle =1(0\cdot(-2)-(-3)\cdot3)-(\lambda-2)(1\cdot(-2)-(-3)\cdot3)+4(1\cdot3-0\cdot3)$
$\displaystyle =1(9)-(\lambda-2)(-2+9)+4(3)$
$\displaystyle =9-7(\lambda-2)+12=0$
$\displaystyle \Rightarrow 21-7\lambda+14=0$
$\displaystyle \Rightarrow 7\lambda=35$
$\displaystyle \Rightarrow \lambda=5$

$\displaystyle \textbf{Question 14: If four points }A,B,C,D\text{ with position vectors }4\widehat{i}+3\widehat{j}+3\widehat{k},\ 5\widehat{i}+x\widehat{j}+7\widehat{k},\ 5\widehat{i}+3\widehat{j}+7\widehat{k}\text{ and }7\widehat{i}+6\widehat{j}+\widehat{k}\text{ respectively are coplanar,} \\ \text{then find the value of }x.\ \text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{OA}=4\widehat{i}+3\widehat{j}+3\widehat{k},\ \overrightarrow{OB}=5\widehat{i}+x\widehat{j}+7\widehat{k},\ \overrightarrow{OC}=5\widehat{i}+3\widehat{j}\text{ and }\overrightarrow{OD}=7\widehat{i}+6\widehat{j}+\widehat{k}.$
$\displaystyle \therefore \overrightarrow{AB}=(5\widehat{i}+x\widehat{j}+7\widehat{k})-(4\widehat{i}+3\widehat{j}+3\widehat{k})=\widehat{i}+(x-3)\widehat{j}+4\widehat{k}$
$\displaystyle \overrightarrow{AC}=(5\widehat{i}+3\widehat{j})-(4\widehat{i}+3\widehat{j}+3\widehat{k})=\widehat{i}-3\widehat{k}$
$\displaystyle \overrightarrow{AD}=(7\widehat{i}+6\widehat{j}+\widehat{k})-(4\widehat{i}+3\widehat{j}+3\widehat{k})=3\widehat{i}+3\widehat{j}-2\widehat{k}$
$\displaystyle \text{Since the given four points are coplanar, the vectors }\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\text{ are also coplanar.}$
$\displaystyle \therefore [\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}]=0$
$\displaystyle \Rightarrow \begin{vmatrix}1&x-3&4\\1&0&-3\\3&3&-2\end{vmatrix}=0$
$\displaystyle \Rightarrow 1(0\cdot(-2)-(-3)\cdot3)-(x-3)(1\cdot(-2)-(-3)\cdot3)+4(1\cdot3-0\cdot3)=0$
$\displaystyle \Rightarrow 1(9)-(x-3)(-2+9)+4(3)=0$
$\displaystyle \Rightarrow 9-7(x-3)+12=0$
$\displaystyle \Rightarrow 9-7x+21+12=0$
$\displaystyle \Rightarrow 7x=42$
$\displaystyle \Rightarrow x=6$
$\displaystyle \text{Thus, the value of }x\text{ is }6.$