Class 12: The Plane – Exercise 29.3

$\displaystyle \textbf{Question 1: }~\text{Find the vector equation of a plane passing through a point having} \\ \text{position vector }2\widehat{i}-\widehat{j}+\widehat{k}\text{ and perpendicular to the vector }4\widehat{i}+2\widehat{j}-3\widehat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}\text{ and }\overrightarrow{n}=4\widehat{i}+2\widehat{j}-3\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}+2\widehat{j}-3\widehat{k}\right)=\left(2\widehat{i}-\widehat{j}+\widehat{k}\right)\cdot\left(4\widehat{i}+2\widehat{j}-3\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}+2\widehat{j}-3\widehat{k}\right)=2\cdot4+(-1)\cdot2+1\cdot(-3)$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}+2\widehat{j}-3\widehat{k}\right)=8-2-3$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}+2\widehat{j}-3\widehat{k}\right)=3$

$\displaystyle \textbf{Question 2: }~\text{Find the cartesian form of equation of a plane whose vector equation is}$
$\displaystyle (i)\ \overrightarrow{r}\cdot(12\widehat{i}-3\widehat{j}+4\widehat{k})+5=0$
$\displaystyle (ii)\ \overrightarrow{r}\cdot(-\widehat{i}+\widehat{j}+2\widehat{k})=9$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the given equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(12\widehat{i}-3\widehat{j}+4\widehat{k}\right)+5=0$
$\displaystyle x\cdot12+y\cdot(-3)+z\cdot4+5=0$
$\displaystyle 12x-3y+4z+5=0$
$\displaystyle \textbf{(ii) }$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the given equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(-\widehat{i}+\widehat{j}+2\widehat{k}\right)=9$
$\displaystyle x\cdot(-1)+y\cdot1+z\cdot2=9$
$\displaystyle -x+y+2z=9$

$\displaystyle \textbf{Question 3: }~\text{Find the vector equations of the coordinate planes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Vector equation of XY-plane}$
$\displaystyle \text{This plane is passing through the origin whose position vector is }\overrightarrow{a}=\overrightarrow{0} \\ \text{ and perpendicular to the z-axis whose position vector is }\widehat{k}.$
$\displaystyle \text{So, the equation of the XY-plane is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \overrightarrow{r}\cdot\widehat{k}=\overrightarrow{0}\cdot\widehat{k}$
$\displaystyle \overrightarrow{r}\cdot\widehat{k}=0$
$\displaystyle \text{Vector equation of YZ-plane}$
$\displaystyle \text{This plane is passing through the origin whose position vector is }\overrightarrow{a}=\overrightarrow{0} \\ \text{ and perpendicular to the x-axis whose position vector is }\widehat{i}.$
$\displaystyle \text{So, the equation of the YZ-plane is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \overrightarrow{r}\cdot\widehat{i}=\overrightarrow{0}\cdot\widehat{i}$
$\displaystyle \overrightarrow{r}\cdot\widehat{i}=0$
$\displaystyle \text{Vector equation of XZ-plane}$
$\displaystyle \text{This plane is passing through the origin whose position vector is }\overrightarrow{a}=\overrightarrow{0} \\ \text{ and perpendicular to the y-axis whose position vector is }\widehat{j}.$
$\displaystyle \text{So, the equation of the XZ-plane is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \overrightarrow{r}\cdot\widehat{j}=\overrightarrow{0}\cdot\widehat{j}$
$\displaystyle \overrightarrow{r}\cdot\widehat{j}=0$

$\displaystyle \textbf{Question 4: }~\text{Find the vector equation of each one of following planes:}$
$\displaystyle (i)\ 2x-y+2z=8$
$\displaystyle (ii)\ x+y-z=5$
$\displaystyle (iii)\ x+y=3$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }$
$\displaystyle \text{The given equation of the plane is}$
$\displaystyle 2x-y+2z=8$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(2\widehat{i}-\widehat{j}+2\widehat{k}\right)=8$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}-\widehat{j}+2\widehat{k}\right)=8\text{, which is the vector equation of the plane.}$
$\displaystyle \text{(Because the vector equation of the plane is }\overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}\text{.)}$
$\displaystyle \textbf{(ii) }$
$\displaystyle \text{The given equation of the plane is}$
$\displaystyle x+y-z=5$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=5$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=5\text{, which is the vector equation of the plane.}$
$\displaystyle \text{(Because the vector equation of the plane is }\overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}\text{.)}$
$\displaystyle \textbf{(iii) }$
$\displaystyle \text{The given equation of the plane is}$
$\displaystyle x+y=3$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(\widehat{i}+\widehat{j}+0\widehat{k}\right)=3$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(\widehat{i}+\widehat{j}\right)=3$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+\widehat{j}\right)=3\text{, which is the vector equation of the plane.}$
$\displaystyle \text{(Because the vector equation of the plane is }\overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}\text{.)}$

$\displaystyle \textbf{Question 5: }~\text{Find the vector and cartesian equations of a plane passing through} \\ \text{the point }(1,-1,1)\text{ and normal to the line joining the points }(1,2,5)\text{ and }(-1,3,1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the given plane passes through the point }(1,-1,1)\text{ and is normal to the line joining }A(1,2,5)\text{ and }B(-1,3,1),$
$\displaystyle \overrightarrow{n}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(-\widehat{i}+3\widehat{j}+\widehat{k}\right)-\left(\widehat{i}+2\widehat{j}+5\widehat{k}\right)=-2\widehat{i}+\widehat{j}-4\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=\widehat{i}-\widehat{j}+\widehat{k}\text{ and }\overrightarrow{n}=-2\widehat{i}+\widehat{j}-4\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(-2\widehat{i}+\widehat{j}-4\widehat{k}\right)=\left(\widehat{i}-\widehat{j}+\widehat{k}\right)\cdot\left(-2\widehat{i}+\widehat{j}-4\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(-2\widehat{i}+\widehat{j}-4\widehat{k}\right)=1\cdot(-2)+(-1)\cdot1+1\cdot(-4)$
$\displaystyle \overrightarrow{r}\cdot\left(-2\widehat{i}+\widehat{j}-4\widehat{k}\right)=-2-1-4$
$\displaystyle \overrightarrow{r}\cdot\left(-2\widehat{i}+\widehat{j}-4\widehat{k}\right)=-7$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}-\widehat{j}+4\widehat{k}\right)=7$
$\displaystyle \text{For Cartesian form, we need to substitute }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation.}$
$\displaystyle \text{Then, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(2\widehat{i}-\widehat{j}+4\widehat{k}\right)=7$
$\displaystyle 2x-y+4z=7$

$\displaystyle \textbf{Question 6: }~\text{If }\overrightarrow{n}\text{ is a vector of magnitude }\sqrt{3} \text{ and is equally} \\ \text{inclined with an acute angle with the coordinate axes, find the vector and cartesian forms} \\ \text{of the equation of a plane which passes through }(2,1,-1)\text{ and is normal to }\overrightarrow{n}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\alpha,\beta\text{ and }\gamma\text{ be the angles made by }\overrightarrow{n}\text{ with }x,y\text{ and }z\text{-axes respectively.}$
$\displaystyle \text{Given that}$
$\displaystyle \alpha=\beta=\gamma$
$\displaystyle \cos\alpha=\cos\beta=\cos\gamma$
$\displaystyle l=m=n\text{, where }l,m,n\text{ are direction cosines of }\overrightarrow{n}.$
$\displaystyle \text{But }l^{2}+m^{2}+n^{2}=1$
$\displaystyle l^{2}+l^{2}+l^{2}=1$
$\displaystyle 3l^{2}=1$
$\displaystyle l^{2}=\frac{1}{3}$
$\displaystyle l=\frac{1}{\sqrt{3}}\text{ (Since }\alpha\text{ is acute, }l=\cos\alpha>0)$
$\displaystyle \text{Thus, }\overrightarrow{n}=\sqrt{3}\left(\frac{1}{\sqrt{3}}\widehat{i}+\frac{1}{\sqrt{3}}\widehat{j}+\frac{1}{\sqrt{3}}\widehat{k}\right)=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k}\text{ and }\overrightarrow{n}=\widehat{i}+\widehat{j}+\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)\cdot\left(\widehat{i}+\widehat{j}+\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=2+1-1$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=2$
$\displaystyle \text{For the Cartesian form, we need to substitute }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation.}$
$\displaystyle \text{Then, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=2$
$\displaystyle x+y+z=2$

$\displaystyle \textbf{Question 7: }~\text{The coordinates of the foot of the perpendicular drawn from the origin} \\ \text{to a plane are }(12,-4,3).\text{ Find the equation of the plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The normal is passing through the points }A(0,0,0)\text{ and }B(12,-4,3)\text{. So,}$
$\displaystyle \overrightarrow{n}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(12\widehat{i}-4\widehat{j}+3\widehat{k}\right)-\left(0\widehat{i}+0\widehat{j}+0\widehat{k}\right)=12\widehat{i}-4\widehat{j}+3\widehat{k}$
$\displaystyle \text{Since the plane passes through }(12,-4,3)\text{, }\overrightarrow{a}=12\widehat{i}-4\widehat{j}+3\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=12\widehat{i}-4\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{n}=12\widehat{i}-4\widehat{j}+3\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(12\widehat{i}-4\widehat{j}+3\widehat{k}\right)=\left(12\widehat{i}-4\widehat{j}+3\widehat{k}\right)\cdot\left(12\widehat{i}-4\widehat{j}+3\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(12\widehat{i}-4\widehat{j}+3\widehat{k}\right)=12^{2}+(-4)^{2}+3^{2}$
$\displaystyle \overrightarrow{r}\cdot\left(12\widehat{i}-4\widehat{j}+3\widehat{k}\right)=144+16+9$
$\displaystyle \overrightarrow{r}\cdot\left(12\widehat{i}-4\widehat{j}+3\widehat{k}\right)=169$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(12\widehat{i}-4\widehat{j}+3\widehat{k}\right)=169$
$\displaystyle 12x-4y+3z=169$

$\displaystyle \textbf{Question 8: }~\text{Find the equation of the plane passing through the point }(2,3,1) \\ \text{ given that the direction ratios of normal to the plane are proportional to }5,3,2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=2\widehat{i}+3\widehat{j}+\widehat{k}\text{ and }\overrightarrow{n}=5\widehat{i}+3\widehat{j}+2\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(5\widehat{i}+3\widehat{j}+2\widehat{k}\right)=\left(2\widehat{i}+3\widehat{j}+\widehat{k}\right)\cdot\left(5\widehat{i}+3\widehat{j}+2\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(5\widehat{i}+3\widehat{j}+2\widehat{k}\right)=2\cdot5+3\cdot3+1\cdot2$
$\displaystyle \overrightarrow{r}\cdot\left(5\widehat{i}+3\widehat{j}+2\widehat{k}\right)=10+9+2$
$\displaystyle \overrightarrow{r}\cdot\left(5\widehat{i}+3\widehat{j}+2\widehat{k}\right)=21$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(5\widehat{i}+3\widehat{j}+2\widehat{k}\right)=21$
$\displaystyle 5x+3y+2z=21$

$\displaystyle \textbf{Question 9: }~\text{If the axes are rectangular and }P\text{ is the point }(2,3,-1), \\ \text{ find the equation of the plane through }P\text{ at right angles to }OP.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The normal is passing through the points }O(0,0,0)\text{ and }P(2,3,-1)\text{. So,}$
$\displaystyle \overrightarrow{n}=\overrightarrow{OP}=\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)-\left(0\widehat{i}+0\widehat{j}+0\widehat{k}\right)=2\widehat{i}+3\widehat{j}-\widehat{k}$
$\displaystyle \text{Since the plane passes through the point }(2,3,-1)\text{, }\overrightarrow{a}=2\widehat{i}+3\widehat{j}-\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=2\widehat{i}+3\widehat{j}-\widehat{k}\text{ and }\overrightarrow{n}=2\widehat{i}+3\widehat{j}-\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=2^{2}+3^{2}+(-1)^{2}$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=4+9+1$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=14$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=14$
$\displaystyle 2x+3y-z=14$

$\displaystyle \textbf{Question 10: }~\text{Find the intercepts made on the coordinate axes by the plane } \\ 2x+y-2z=3\text{ and find also the direction cosines of the normal to the plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation of the plane is }2x+y-2z=3$
$\displaystyle \text{Dividing both sides by }3\text{, we get}$
$\displaystyle \frac{2x}{3}+\frac{y}{3}+\frac{-2z}{3}=\frac{3}{3}$
$\displaystyle \frac{x}{\frac{3}{2}}+\frac{y}{3}+\frac{z}{\frac{-3}{2}}=1\quad (1)$
$\displaystyle \text{We know that the equation of the plane whose intercepts on the coordinate axes are}$
$\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\quad (2)$
$\displaystyle \text{Comparing (1) and (2), we get}$
$\displaystyle a=\frac{3}{2},\ b=3,\ c=\frac{-3}{2}$
$\displaystyle \text{Finding the direction cosines of the normal}$
$\displaystyle \text{The given equation of the plane is }2x+y-2z=8$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(2\widehat{i}+\widehat{j}-2\widehat{k}\right)=8$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+\widehat{j}-2\widehat{k}\right)=8\text{, which is the vector equation of the plane (because }\overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}\text{).}$
$\displaystyle \text{where the normal to the plane is }\overrightarrow{n}=2\widehat{i}+\widehat{j}-2\widehat{k}$
$\displaystyle \left|\overrightarrow{n}\right|=\sqrt{4+1+4}=3$
$\displaystyle \text{So, the unit vector perpendicular to }\overrightarrow{n}\text{ is }\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{2\widehat{i}+\widehat{j}-2\widehat{k}}{3}=\frac{2}{3}\widehat{i}+\frac{1}{3}\widehat{j}-\frac{2}{3}\widehat{k}$
$\displaystyle \text{So, the direction cosines of the normal to the plane are }\frac{2}{3},\frac{1}{3},\frac{-2}{3}$

$\displaystyle \textbf{Question 11: }~\text{A plane passes through the point }(1,-2,5)\text{ and is perpendicular} \\ \text{to the line joining the origin to the point }3\widehat{i}+\widehat{j}-\widehat{k}.\text{ Find the vector and cartesian} \\ \text{forms of the } \text{equation of the plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The normal is passing through the points }A(0,0,0)\text{ and }B(3,1,-1)\text{. So,}$
$\displaystyle \overrightarrow{n}=\overrightarrow{OB}=\left(3\widehat{i}+\widehat{j}-\widehat{k}\right)-\left(0\widehat{i}+0\widehat{j}+0\widehat{k}\right)=3\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \text{Since the plane passes through }(1,-2,5)\text{, }\overrightarrow{a}=\widehat{i}-2\widehat{j}+5\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=\widehat{i}-2\widehat{j}+5\widehat{k}\text{ and }\overrightarrow{n}=3\widehat{i}+\widehat{j}-\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(3\widehat{i}+\widehat{j}-\widehat{k}\right)=\left(\widehat{i}-2\widehat{j}+5\widehat{k}\right)\cdot\left(3\widehat{i}+\widehat{j}-\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(3\widehat{i}+\widehat{j}-\widehat{k}\right)=1\cdot3+(-2)\cdot1+5\cdot(-1)$
$\displaystyle \overrightarrow{r}\cdot\left(3\widehat{i}+\widehat{j}-\widehat{k}\right)=3-2-5$
$\displaystyle \overrightarrow{r}\cdot\left(3\widehat{i}+\widehat{j}-\widehat{k}\right)=-4$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(3\widehat{i}+\widehat{j}-\widehat{k}\right)=-4$
$\displaystyle 3x+y-z=-4$

$\displaystyle \textbf{Question 12: }~\text{Find the equation of the plane that bisects the line segment joining} \\ \text{points }(1,2,3)\text{ and }(3,4,5)\text{ and is at right angle to it.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The normal is passing through the points }A(1,2,3)\text{ and }B(3,4,5)\text{. So,}$
$\displaystyle \overrightarrow{n}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(3\widehat{i}+4\widehat{j}+5\widehat{k}\right)-\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)=2\widehat{i}+2\widehat{j}+2\widehat{k}$
$\displaystyle \text{Mid-point of }AB=\left(\frac{1+3}{2},\frac{2+4}{2},\frac{3+5}{2}\right)=(2,3,4)$
$\displaystyle \text{Since the plane passes through }(2,3,4)\text{, }\overrightarrow{a}=2\widehat{i}+3\widehat{j}+4\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=2\widehat{i}+3\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{n}=2\widehat{i}+2\widehat{j}+2\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+2\widehat{j}+2\widehat{k}\right)=\left(2\widehat{i}+3\widehat{j}+4\widehat{k}\right)\cdot\left(2\widehat{i}+2\widehat{j}+2\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+2\widehat{j}+2\widehat{k}\right)=2\cdot2+3\cdot2+4\cdot2$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+2\widehat{j}+2\widehat{k}\right)=4+6+8$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+2\widehat{j}+2\widehat{k}\right)=18$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=9$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=9$
$\displaystyle x+y+z=9$

$\displaystyle \textbf{Question 13: }~\text{Show that the normals to the following pairs of planes are} \\ \text{perpendicular to each other:}$
$\displaystyle (i)\ x-y+z-2=0\text{ and }3x+2y-z+4=0$
$\displaystyle (ii)\ \overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+3\widehat{k})=5\text{ and }\overrightarrow{r}\cdot(2\widehat{i}-2\widehat{j}-2\widehat{k})=5$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }$
$\displaystyle \text{Let }\overrightarrow{n}_{1}\text{ and }\overrightarrow{n}_{2}\text{ be the vectors which are normals to the planes }x-y+z=2\text{ and }3x+2y-z=-4\text{ respectively.}$
$\displaystyle \text{The given equations of the planes are}$
$\displaystyle x-y+z=2;\ 3x+2y-z=-4$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(\widehat{i}-\widehat{j}+\widehat{k}\right)=2;\ \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(3\widehat{i}+2\widehat{j}-\widehat{k}\right)=-4$
$\displaystyle \overrightarrow{n}_{1}=\widehat{i}-\widehat{j}+\widehat{k};\ \overrightarrow{n}_{2}=3\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{n}_{1}\cdot\overrightarrow{n}_{2}=\left(\widehat{i}-\widehat{j}+\widehat{k}\right)\cdot\left(3\widehat{i}+2\widehat{j}-\widehat{k}\right)$
$\displaystyle \overrightarrow{n}_{1}\cdot\overrightarrow{n}_{2}=1\cdot3+(-1)\cdot2+1\cdot(-1)$
$\displaystyle \overrightarrow{n}_{1}\cdot\overrightarrow{n}_{2}=3-2-1=0$
$\displaystyle \text{So, the normals to the given planes are perpendicular to each other.}$
$\displaystyle \textbf{(ii) }$
$\displaystyle \text{Let }\overrightarrow{n}_{1}\text{ and }\overrightarrow{n}_{2}\text{ be the vectors which are normals to the planes }\overrightarrow{r}\cdot\left(2\widehat{i}-\widehat{j}+3\widehat{k}\right)=5\text{ and }\overrightarrow{r}\cdot\left(2\widehat{i}-2\widehat{j}-2\widehat{k}\right)=5\text{ respectively.}$
$\displaystyle \text{The given equations of the planes are}$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}-\widehat{j}+3\widehat{k}\right)=5;\ \overrightarrow{r}\cdot\left(2\widehat{i}-2\widehat{j}-2\widehat{k}\right)=5$
$\displaystyle \overrightarrow{n}_{1}=2\widehat{i}-\widehat{j}+3\widehat{k};\ \overrightarrow{n}_{2}=2\widehat{i}-2\widehat{j}-2\widehat{k}$
$\displaystyle \text{Now, }\overrightarrow{n}_{1}\cdot\overrightarrow{n}_{2}=\left(2\widehat{i}-\widehat{j}+3\widehat{k}\right)\cdot\left(2\widehat{i}-2\widehat{j}-2\widehat{k}\right)$
$\displaystyle \overrightarrow{n}_{1}\cdot\overrightarrow{n}_{2}=2\cdot2+(-1)\cdot(-2)+3\cdot(-2)$
$\displaystyle \overrightarrow{n}_{1}\cdot\overrightarrow{n}_{2}=4+2-6=0$
$\displaystyle \text{So, the normals to the given planes are perpendicular to each other.}$

$\displaystyle \textbf{Question 14: }~\text{Show that the normal vector to the plane }2x+2y+2z=3 \\ \text{ is equally inclined with the coordinate axes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation of the plane is}$
$\displaystyle 2x+2y+2z=8$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(2\widehat{i}+2\widehat{j}+2\widehat{k}\right)=8$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+2\widehat{j}+2\widehat{k}\right)=8\text{, which is the vector equation of the plane.}$
$\displaystyle \text{(Because the vector equation of the plane is }\overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}\text{,}$
$\displaystyle \text{where the normal to the plane is }\overrightarrow{n}=2\widehat{i}+2\widehat{j}+2\widehat{k}\text{.)}$
$\displaystyle \left|\overrightarrow{n}\right|=\sqrt{4+4+4}=2\sqrt{3}$
$\displaystyle \text{So, the unit vector perpendicular to }\overrightarrow{n}\text{ is }\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{2\widehat{i}+2\widehat{j}+2\widehat{k}}{2\sqrt{3}}=\frac{1}{\sqrt{3}}\widehat{i}+\frac{1}{\sqrt{3}}\widehat{j}+\frac{1}{\sqrt{3}}\widehat{k}$
$\displaystyle \text{So, the direction cosines of the normal to the plane are }l=\frac{1}{\sqrt{3}},\ m=\frac{1}{\sqrt{3}},\ n=\frac{1}{\sqrt{3}}$
$\displaystyle \text{Let }\alpha,\beta\text{ and }\gamma\text{ be the angles made by the given plane with the coordinate axes.}$
$\displaystyle \text{Then,}$
$\displaystyle l=\cos\alpha=\frac{1}{\sqrt{3}},\ m=\cos\beta=\frac{1}{\sqrt{3}},\ n=\cos\gamma=\frac{1}{\sqrt{3}}$
$\displaystyle \cos\alpha=\cos\beta=\cos\gamma$
$\displaystyle \alpha=\beta=\gamma$
$\displaystyle \text{So, the given plane is equally inclined to the coordinate axes.}$

$\displaystyle \textbf{Question 15: }~\text{Find a vector of magnitude }26\text{ units normal to the plane } \\ 12x-3y+4z=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation of the plane is}$
$\displaystyle 12x-3y+4z=1$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(12\widehat{i}-3\widehat{j}+4\widehat{k}\right)=1$
$\displaystyle \overrightarrow{r}\cdot\left(12\widehat{i}-3\widehat{j}+4\widehat{k}\right)=1\text{, which is the vector equation of the plane.}$
$\displaystyle \text{Because the vector equation of the plane is }\overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}\text{)}$
$\displaystyle \text{So, the normal vector is }\overrightarrow{n}=12\widehat{i}-3\widehat{j}+4\widehat{k}$
$\displaystyle \left|\overrightarrow{n}\right|=\sqrt{144+9+16}=13$
$\displaystyle \text{Unit vector parallel to }\overrightarrow{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{12\widehat{i}-3\widehat{j}+4\widehat{k}}{13}$
$\displaystyle \text{So, the vector of magnitude }26\text{ units normal to the plane}$
$\displaystyle =26\times\frac{12\widehat{i}-3\widehat{j}+4\widehat{k}}{13}$
$\displaystyle =2\left(12\widehat{i}-3\widehat{j}+4\widehat{k}\right)$
$\displaystyle =24\widehat{i}-6\widehat{j}+8\widehat{k}$

$\displaystyle \textbf{Question 16: }~\text{If the line drawn from }(4,-1,2)\text{ meets a plane at right angles} \\ \text{at the point }(-10,5,4),\text{ find the equation of the plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The normal is passing through the points }A(4,-1,2)\text{ and }B(-10,5,4)\text{. So,}$
$\displaystyle \overrightarrow{n}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(-10\widehat{i}+5\widehat{j}+4\widehat{k}\right)-\left(4\widehat{i}-\widehat{j}+2\widehat{k}\right)=-14\widehat{i}+6\widehat{j}+2\widehat{k}$
$\displaystyle \text{Since the plane passes through }(-10,5,4)\text{, }\overrightarrow{a}=-10\widehat{i}+5\widehat{j}+4\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=-10\widehat{i}+5\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{n}=-14\widehat{i}+6\widehat{j}+2\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(-14\widehat{i}+6\widehat{j}+2\widehat{k}\right)=\left(-10\widehat{i}+5\widehat{j}+4\widehat{k}\right)\cdot\left(-14\widehat{i}+6\widehat{j}+2\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(-14\widehat{i}+6\widehat{j}+2\widehat{k}\right)=(-10)(-14)+5\cdot6+4\cdot2$
$\displaystyle \overrightarrow{r}\cdot\left(-14\widehat{i}+6\widehat{j}+2\widehat{k}\right)=140+30+8$
$\displaystyle \overrightarrow{r}\cdot\left(-14\widehat{i}+6\widehat{j}+2\widehat{k}\right)=178$
$\displaystyle \overrightarrow{r}\cdot\left(7\widehat{i}-3\widehat{j}-\widehat{k}\right)=-89$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(7\widehat{i}-3\widehat{j}-\widehat{k}\right)=-89$
$\displaystyle 7x-3y-z=-89$
$\displaystyle 7x-3y-z+89=0$

$\displaystyle \textbf{Question 17: }~\text{Find the equation of the plane which bisects the line segment} \\ \text{joining the points }(-1,2,3)\text{ and }(3,-5,6)\text{ at right angles.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The normal is passing through the points }A(-1,2,3)\text{ and }B(3,-5,6)\text{. So,}$
$\displaystyle \overrightarrow{n}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(3\widehat{i}-5\widehat{j}+6\widehat{k}\right)-\left(-\widehat{i}+2\widehat{j}+3\widehat{k}\right)=4\widehat{i}-7\widehat{j}+3\widehat{k}$
$\displaystyle \text{Mid-point of }AB=\left(\frac{-1+3}{2},\frac{2-5}{2},\frac{3+6}{2}\right)=\left(1,\frac{-3}{2},\frac{9}{2}\right)$
$\displaystyle \text{Since the plane passes through }\left(1,\frac{-3}{2},\frac{9}{2}\right)\text{, }\overrightarrow{a}=\widehat{i}-\frac{3}{2}\widehat{j}+\frac{9}{2}\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=\widehat{i}-\frac{3}{2}\widehat{j}+\frac{9}{2}\widehat{k}\text{ and }\overrightarrow{n}=4\widehat{i}-7\widehat{j}+3\widehat{k}\text{, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}-7\widehat{j}+3\widehat{k}\right)=\left(\widehat{i}-\frac{3}{2}\widehat{j}+\frac{9}{2}\widehat{k}\right)\cdot\left(4\widehat{i}-7\widehat{j}+3\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}-7\widehat{j}+3\widehat{k}\right)=1\cdot4+\left(-\frac{3}{2}\right)(-7)+\frac{9}{2}\cdot3$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}-7\widehat{j}+3\widehat{k}\right)=4+\frac{21}{2}+\frac{27}{2}$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}-7\widehat{j}+3\widehat{k}\right)=28$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(4\widehat{i}-7\widehat{j}+3\widehat{k}\right)=28$
$\displaystyle 4x-7y+3z=28$
$\displaystyle 4x-7y+3z-28=0$

$\displaystyle \textbf{Question 18: }~\text{Find the vector and cartesian equationsnd equations of the plane} \\ \text{which passes through the point }(5,2,-4)\text{ and perpendicular to the line} \\ \text{with direction ratios }2,3,-1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=5\widehat{i}+2\widehat{j}-4\widehat{k}\text{ and }\overrightarrow{n}=2\widehat{i}+3\widehat{j}-\widehat{k}\text{ (because the direction ratios of }\overrightarrow{n}\text{ are }2,3,-1),\text{ we get}$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=\left(5\widehat{i}+2\widehat{j}-4\widehat{k}\right)\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=5\cdot2+2\cdot3+(-4)\cdot(-1)$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=10+6+4$
$\displaystyle \overrightarrow{r}\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=20$
$\displaystyle \text{For Cartesian form, we need to substitute }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in this equation. Then, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)=20$
$\displaystyle 2x+3y-z=20$

$\displaystyle \textbf{Question 19: }~\text{If }O\text{ be the origin and the coordinates of }P\text{ be }(1,2,-3), \\\text{ then find the equation of the plane passing through }P\text{ and perpendicular to }OP.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The normal is passing through the points }O(0,0,0)\text{ and }P(1,2,-3)\text{. So,}$
$\displaystyle \overrightarrow{n}=\overrightarrow{OP}=\left(\widehat{i}+2\widehat{j}-3\widehat{k}\right)-\left(0\widehat{i}+0\widehat{j}+0\widehat{k}\right)=\widehat{i}+2\widehat{j}-3\widehat{k}$
$\displaystyle \text{Since the plane passes through }P(1,2,-3)\text{, }\overrightarrow{a}=\widehat{i}+2\widehat{j}-3\widehat{k}$
$\displaystyle \text{We know that the vector equation of the plane passing through a point }\overrightarrow{a}\text{ and normal to }\overrightarrow{n}\text{ is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \text{Substituting }\overrightarrow{a}=\widehat{i}+2\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{n}=\widehat{i}+2\widehat{j}-3\widehat{k}\text{ in the relation, we get}$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+2\widehat{j}-3\widehat{k}\right)=\left(\widehat{i}+2\widehat{j}-3\widehat{k}\right)\cdot\left(\widehat{i}+2\widehat{j}-3\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+2\widehat{j}-3\widehat{k}\right)=1^{2}+2^{2}+(-3)^{2}$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+2\widehat{j}-3\widehat{k}\right)=1+4+9$
$\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+2\widehat{j}-3\widehat{k}\right)=14$
$\displaystyle \text{Substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in the vector equation, we get}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(\widehat{i}+2\widehat{j}-3\widehat{k}\right)=14$
$\displaystyle x+2y-3z=14$

$\displaystyle \textbf{Question 20: }~\text{If }O\text{ is the origin and the coordinates of }A\text{ are }(a,b,c). \\\text{ Find the direction cosines of }OA\text{ and the equation of the plane through }A \\ \text{ at right angles to }OA.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that }O\text{ is the origin and the coordinates of }A\text{ are }(a,b,c).$
$\displaystyle \text{The direction ratios of }OA\text{ are proportional to}$
$\displaystyle a-0,\ b-0,\ c-0$
$\displaystyle \text{Direction cosines of }OA\text{ are}$
$\displaystyle \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}},\ \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},\ \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$\displaystyle \text{The normal vector to the required plane is}$
$\displaystyle a\widehat{i}+b\widehat{j}+c\widehat{k}$
$\displaystyle \text{The vector equation of the plane through }A(a,b,c)\text{ and perpendicular to }OA\text{ is}$
$\displaystyle \overrightarrow{r}-\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)\cdot\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0\ \left(\overrightarrow{r}-\overrightarrow{a}\right)\cdot\overrightarrow{n}=0$
$\displaystyle \overrightarrow{r}\cdot\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)\cdot\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)$
$\displaystyle \overrightarrow{r}\cdot\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=a^{2}+b^{2}+c^{2}$
$\displaystyle \text{The Cartesian equation of this plane is}$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=a^{2}+b^{2}+c^{2}$
$\displaystyle ax+by+cz=a^{2}+b^{2}+c^{2}$

$\displaystyle \textbf{Question 21: }~\text{Find the vector equation of the plane with intercepts }3,\ -4\text{ and } \\ 2\text{ on }x,y\text{ and }z\text{ axes respectively. \ [CBSE\ 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane in the intercept form is}$
$\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\text{ where }a,b\text{ and }c\text{ are the intercepts on the }x,y\text{ and }z\text{-axes respectively.}$
$\displaystyle \text{It is given that the intercepts made by the plane on the }x,y\text{ and }z\text{-axes are }3,-4\text{ and }2\text{ respectively.}$
$\displaystyle a=3,\ b=-4,\ c=2$
$\displaystyle \text{Thus, the equation of the plane is}$
$\displaystyle \frac{x}{3}+\frac{y}{-4}+\frac{z}{2}=1$
$\displaystyle 4x-3y+6z=12$
$\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(4\widehat{i}-3\widehat{j}+6\widehat{k}\right)=12$
$\displaystyle \overrightarrow{r}\cdot\left(4\widehat{i}-3\widehat{j}+6\widehat{k}\right)=12$
$\displaystyle \text{This is the vector form of the equation of the given plane.}$