\displaystyle \textbf{Question 1: }~\text{Find the vector equation of the plane passing through the points } \\ (1,1,1),\ (1,-1,1)\text{ and }(-7,-3,-5).
\displaystyle \text{Answer:}
\displaystyle \text{Let } A(1,1,1),\ B(1,-1,1) \text{ and } C(-7,-3,-5) \text{ be the coordinates.}
\displaystyle \text{The required plane passes through the point } A(1,1,1) \text{ whose position vector is } \overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k} \text{ and is normal to the vector } \overrightarrow{n} \text{ given by } \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}.
\displaystyle \text{Clearly, } \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(\widehat{i}-\widehat{j}+\widehat{k})-(\widehat{i}+\widehat{j}+\widehat{k})=0\widehat{i}-2\widehat{j}+0\widehat{k}.
\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(-7\widehat{i}-3\widehat{j}-5\widehat{k})-(\widehat{i}+\widehat{j}+\widehat{k})=-8\widehat{i}-4\widehat{j}-6\widehat{k}.
\displaystyle \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\0&-2&0\\-8&-4&-6\end{vmatrix}
\displaystyle =\widehat{i}\big((-2)(-6)-0(-4)\big)-\widehat{j}\big(0(-6)-0(-8)\big)+\widehat{k}\big(0(-4)-(-2)(-8)\big)
\displaystyle =12\widehat{i}+0\widehat{j}-16\widehat{k}.
\displaystyle \text{The vector equation of the required plane is } \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(12\widehat{i}+0\widehat{j}-16\widehat{k})=(\widehat{i}+\widehat{j}+\widehat{k})\cdot(12\widehat{i}+0\widehat{j}-16\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot(12\widehat{i}-16\widehat{k})=12+0-16.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(12\widehat{i}-16\widehat{k})=-4.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(3\widehat{i}-4\widehat{k})=-1.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(3\widehat{i}-4\widehat{k})+1=0.

\displaystyle \textbf{Question 2: }~\text{Find the vector equation of the plane passing through the points } \\ P(2,5,-3),\ Q(-2,-3,5)\text{ and }R(5,3,-3). 
\displaystyle \text{Answer:}
\displaystyle \text{The required plane passes through the point } P(2,5,-3) \text{ whose position vector is } \overrightarrow{a}=2\widehat{i}+5\widehat{j}-3\widehat{k} \text{ and is normal to the vector } \overrightarrow{n} \text{ given by } \overrightarrow{n}=\overrightarrow{PQ}\times\overrightarrow{PR}.
\displaystyle \text{Clearly, } \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=(-2\widehat{i}-3\widehat{j}+5\widehat{k})-(2\widehat{i}+5\widehat{j}-3\widehat{k})=-4\widehat{i}-8\widehat{j}+8\widehat{k}.
\displaystyle \overrightarrow{PR}=\overrightarrow{OR}-\overrightarrow{OP}=(5\widehat{i}+3\widehat{j}-3\widehat{k})-(2\widehat{i}+5\widehat{j}-3\widehat{k})=3\widehat{i}-2\widehat{j}-0\widehat{k}.
\displaystyle \overrightarrow{n}=\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-4&-8&8\\3&-2&0\end{vmatrix}
\displaystyle =\widehat{i}\big((-8)(0)-8(-2)\big)-\widehat{j}\big((-4)(0)-8(3)\big)+\widehat{k}\big((-4)(-2)-(-8)(3)\big)
\displaystyle =16\widehat{i}+24\widehat{j}+32\widehat{k}.
\displaystyle \text{The vector equation of the required plane is } \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(16\widehat{i}+24\widehat{j}+32\widehat{k})=(2\widehat{i}+5\widehat{j}-3\widehat{k})\cdot(16\widehat{i}+24\widehat{j}+32\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot(16\widehat{i}+24\widehat{j}+32\widehat{k})=32+120-96.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(16\widehat{i}+24\widehat{j}+32\widehat{k})=56.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(2\widehat{i}+3\widehat{j}+4\widehat{k})=7.

\displaystyle \textbf{Question 3: }~\text{Find the vector equation of the plane passing through points } \\ A(a,0,0),\ B(0,b,0)\text{ and }C(0,0,c). \text{ Reduce it to normal form. If plane } \\ ABC\text{ is at a distance }p\text{ from the origin, prove that }\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}.
\displaystyle \text{Answer:}
\displaystyle \text{The required plane passes through the point } A(a,0,0) \text{ whose position vector is } \overrightarrow{a}=a\widehat{i}+0\widehat{j}+0\widehat{k} \text{ and is normal to the vector } \overrightarrow{n} \text{ given by } \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}.
\displaystyle \text{Clearly, } \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(0\widehat{i}+b\widehat{j}+0\widehat{k})-(a\widehat{i}+0\widehat{j}+0\widehat{k})=-a\widehat{i}+b\widehat{j}+0\widehat{k}.
\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(0\widehat{i}+0\widehat{j}+c\widehat{k})-(a\widehat{i}+0\widehat{j}+0\widehat{k})=-a\widehat{i}+0\widehat{j}+c\widehat{k}.
\displaystyle \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-a&b&0\\-a&0&c\end{vmatrix}
\displaystyle =\widehat{i}\big(bc-0\big)-\widehat{j}\big(-ac-0\big)+\widehat{k}\big(0-ab\big)
\displaystyle =bc\,\widehat{i}+ac\,\widehat{j}+ab\,\widehat{k}.
\displaystyle \text{The vector equation of the required plane is } \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(bc\,\widehat{i}+ac\,\widehat{j}+ab\,\widehat{k})=(a\widehat{i}+0\widehat{j}+0\widehat{k})\cdot(bc\,\widehat{i}+ac\,\widehat{j}+ab\,\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot(bc\,\widehat{i}+ac\,\widehat{j}+ab\,\widehat{k})=abc+0+0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(bc\,\widehat{i}+ac\,\widehat{j}+ab\,\widehat{k})=abc.\qquad(1)
\displaystyle \text{Now, } \lvert\overrightarrow{n}\rvert=\sqrt{(bc)^2+(ac)^2+(ab)^2}=\sqrt{b^2c^2+a^2c^2+a^2b^2}.
\displaystyle \text{For reducing (1) to normal form, we divide both sides by } \sqrt{b^2c^2+a^2c^2+a^2b^2}. \text{ Then we get}
\displaystyle \overrightarrow{r}\cdot\left(\frac{bc\,\widehat{i}+ac\,\widehat{j}+ab\,\widehat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\right)=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}, \text{ which is the normal form of plane (1).}
\displaystyle \text{So, the distance of plane (1) from the origin is}
\displaystyle p=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}.
\displaystyle \Rightarrow \frac{1}{p}=\frac{\sqrt{b^2c^2+a^2c^2+a^2b^2}}{abc}.
\displaystyle \Rightarrow \frac{1}{p^2}=\frac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}.
\displaystyle \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}.

\displaystyle \textbf{Question 4: }~\text{Find the vector equation of the plane passing through the points } \\ (1,1,-1),\ (6,4,-5)\text{ and }(-4,-2,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let } A(1,1,-1),\ B(6,4,-5) \text{ and } C(-4,-2,3).
\displaystyle \text{The required plane passes through the point } A(1,1,-1) \text{ whose position vector is } \overrightarrow{a}=\widehat{i}+\widehat{j}-\widehat{k} \text{ and is normal to the vector } \overrightarrow{n} \text{ given by } \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}.
\displaystyle \text{Clearly, } \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(6\widehat{i}+4\widehat{j}-5\widehat{k})-(\widehat{i}+\widehat{j}-\widehat{k})=5\widehat{i}+3\widehat{j}-4\widehat{k}.
\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(-4\widehat{i}-2\widehat{j}+3\widehat{k})-(\widehat{i}+\widehat{j}-\widehat{k})=-5\widehat{i}-3\widehat{j}+4\widehat{k}.
\displaystyle \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\5&3&-4\\-5&-3&4\end{vmatrix}
\displaystyle =\widehat{i}\big(3\cdot4-(-4)(-3)\big)-\widehat{j}\big(5\cdot4-(-4)(-5)\big)+\widehat{k}\big(5(-3)-3(-5)\big)
\displaystyle =0\widehat{i}+0\widehat{j}+0\widehat{k}=\overrightarrow{0}.
\displaystyle \text{So, the given points are collinear.}
\displaystyle \text{Thus, there will be infinitely many planes passing through these points.}
\displaystyle \text{Their equations (passing through } (1,1,-1)) \text{ are given by}
\displaystyle a(x-1)+b(y-1)+c(z+1)=0.\qquad(1)
\displaystyle \text{Since this passes through } B(6,4,-5),
\displaystyle a(6-1)+b(4-1)+c(-5+1)=0.
\displaystyle \Rightarrow 5a+3b-4c=0.\qquad(2)
\displaystyle \text{From (1) and (2), the equations of the infinite planes are}
\displaystyle a(x-1)+b(y-1)+c(z+1)=0,\ \text{where } 5a+3b-4c=0.

\displaystyle \textbf{Question 5: }~\text{Find the vector equation of the plane passing through the points } \\ 3\widehat{i}+4\widehat{j}+2\widehat{k},\ 2\widehat{i}-2\widehat{j}-\widehat{k}\text{ and }7\widehat{i}+6\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A(3,4,2),\ B(2,-2,-1) \text{ and } C(7,0,6) \text{ be the points represented by the given position vectors.}
\displaystyle \text{The required plane passes through the point } A(3,4,2) \text{ whose position vector is } \overrightarrow{a}=3\widehat{i}+4\widehat{j}+2\widehat{k} \text{ and is normal to the vector } \overrightarrow{n} \text{ given by } \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}.
\displaystyle \text{Clearly, } \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(2\widehat{i}-2\widehat{j}-\widehat{k})-(3\widehat{i}+4\widehat{j}+2\widehat{k})=-\widehat{i}-6\widehat{j}-3\widehat{k}.
\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(7\widehat{i}+0\widehat{j}+6\widehat{k})-(3\widehat{i}+4\widehat{j}+2\widehat{k})=4\widehat{i}-4\widehat{j}+4\widehat{k}.
\displaystyle \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&-6&-3\\4&-4&4\end{vmatrix}
\displaystyle =\widehat{i}\big((-6)(4)-(-3)(-4)\big)-\widehat{j}\big((-1)(4)-(-3)(4)\big)+\widehat{k}\big((-1)(-4)-(-6)(4)\big)
\displaystyle =-36\widehat{i}-8\widehat{j}+28\widehat{k}.
\displaystyle \text{The vector equation of the required plane is } \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-36\widehat{i}-8\widehat{j}+28\widehat{k})=(3\widehat{i}+4\widehat{j}+2\widehat{k})\cdot(-36\widehat{i}-8\widehat{j}+28\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-4(9\widehat{i}+2\widehat{j}-7\widehat{k})\big)=-108-32+56.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-4(9\widehat{i}+2\widehat{j}-7\widehat{k})\big)=-84.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(9\widehat{i}+2\widehat{j}-7\widehat{k})=21.

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