Class 12: The Plane – Exercise 29.11

$\displaystyle \textbf{Question 1: }~\text{Find the angle between the line } \\ \overrightarrow{r}=(2\widehat{i}+3\widehat{j}+9\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+4\widehat{k})\text{ and the plane }\overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=5.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the angle }\theta\text{ between the line }\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\text{ and the plane }\overrightarrow{r}\cdot \overrightarrow{n}=d\text{ is given by}$
$\displaystyle \sin\theta=\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$
$\displaystyle \text{Here,}$
$\displaystyle \overrightarrow{b}=2\widehat{i}+3\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{n}=\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{So , }\sin\theta=\frac{(2\widehat{i}+3\widehat{j}+4\widehat{k})\cdot(\widehat{i}+\widehat{j}+\widehat{k})}{\left|2\widehat{i}+3\widehat{j}+4\widehat{k}\right|\left|\widehat{i}+\widehat{j}+\widehat{k}\right|}$
$\displaystyle =\frac{2+3+4}{\sqrt{4+9+16}\sqrt{1+1+1}}$
$\displaystyle =\frac{9}{\sqrt{29}\sqrt{3}}$
$\displaystyle =\frac{3\sqrt{3}}{\sqrt{29}}$
$\displaystyle \Rightarrow \theta=\sin^{-1}\!\left(\frac{3\sqrt{3}}{\sqrt{29}}\right)$

$\displaystyle \textbf{Question 2: }~\text{Find the angle between the line }\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{1} \\ \text{ and the plane }2x+y-z=4.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given line is parallel to the vector }\overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}\text{ and the given plane is normal to the vector }\overrightarrow{n}=2\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \text{We know that the angle }\theta\text{ between the line and the plane is given by}$
$\displaystyle \sin\theta=\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{(\widehat{i}-\widehat{j}+\widehat{k})\cdot(2\widehat{i}+\widehat{j}-\widehat{k})}{\left|\widehat{i}-\widehat{j}+\widehat{k}\right|\left|2\widehat{i}+\widehat{j}-\widehat{k}\right|}$
$\displaystyle =\frac{2-1-1}{\sqrt{1+1+1}\sqrt{4+1+1}}$
$\displaystyle =\frac{0}{\sqrt{3}\sqrt{6}}$
$\displaystyle =0$
$\displaystyle \Rightarrow \theta=\sin^{-1}(0)=0$

$\displaystyle \textbf{Question 3: }~\text{Find the angle between the line joining the points }(3,-4,-2)\text{ and } \\ (12,2,0)\text{ and the plane }3x-y+z=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{It is given that the line passes through }A(3,-4,-2)\text{ and }B(12,2,0)$
$\displaystyle \text{So, }\overrightarrow{b}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=12\widehat{i}+2\widehat{j}+0\widehat{k}-(3\widehat{i}-4\widehat{j}-2\widehat{k})$
$\displaystyle =9\widehat{i}+6\widehat{j}+2\widehat{k}$
$\displaystyle \text{The given line is parallel to the vector }\overrightarrow{b}=9\widehat{i}+6\widehat{j}+2\widehat{k}\text{ and the given plane is normal to the vector }\overrightarrow{n}=3\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \text{We know that the angle }\theta\text{ between the line and the plane is given by}$
$\displaystyle \sin\theta=\frac{\overrightarrow{b}\cdot\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{(9\widehat{i}+6\widehat{j}+2\widehat{k})\cdot(3\widehat{i}-\widehat{j}+\widehat{k})}{\left|9\widehat{i}+6\widehat{j}+2\widehat{k}\right|\left|3\widehat{i}-\widehat{j}+\widehat{k}\right|}$
$\displaystyle =\frac{27-6+2}{\sqrt{81+36+4}\sqrt{9+1+1}}$
$\displaystyle =\frac{23}{\sqrt{121}\sqrt{11}}$
$\displaystyle =\frac{23}{11\sqrt{11}}$
$\displaystyle \Rightarrow \theta=\sin^{-1}\!\left(\frac{23}{11\sqrt{11}}\right)$

$\displaystyle \textbf{Question 4: }~\text{The line }\overrightarrow{r}=\widehat{i}+\lambda(2\widehat{i}-m\widehat{j}-3\widehat{k})\text{ is parallel to the plane } \\ \overrightarrow{r}\cdot(m\widehat{i}+3\widehat{j}+\widehat{k})=4.\text{ Find }m.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given line is parallel to the vector }\overrightarrow{b}=2\widehat{i}-m\widehat{j}-3\widehat{k}\text{ and the given plane is normal to the vector }\overrightarrow{n}=m\widehat{i}+3\widehat{j}+\widehat{k}$
$\displaystyle \text{If the line is parallel to the plane, the normal to the plane is perpendicular to the line.}$
$\displaystyle \Rightarrow \overrightarrow{b}\perp\overrightarrow{n}$
$\displaystyle \Rightarrow \overrightarrow{b}\cdot\overrightarrow{n}=0$
$\displaystyle \Rightarrow (2\widehat{i}-m\widehat{j}-3\widehat{k})\cdot(m\widehat{i}+3\widehat{j}+\widehat{k})=0$
$\displaystyle \Rightarrow 2m-3m-3=0$
$\displaystyle \Rightarrow -m-3=0$
$\displaystyle \Rightarrow m=-3$

$\displaystyle \textbf{Question 5: }~\text{Show that the line whose vector equation is }\overrightarrow{r}=2\widehat{i}+5\widehat{j}+7\widehat{k}+\lambda(\widehat{i}+3\widehat{j}+4\widehat{k})\text{ is parallel to the plane whose vector equation is } \\ \overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=7.\text{ Also, find the distance between them.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given plane passes through the point with position vector }\overrightarrow{a}=2\widehat{i}+5\widehat{j}+7\widehat{k}\text{ and is parallel to the vector }\overrightarrow{b}=\widehat{i}+3\widehat{j}+4\widehat{k}$
$\displaystyle \text{The given plane is }\overrightarrow{r}\cdot(\widehat{i}+\widehat{j}-\widehat{k})=7$
$\displaystyle \text{So, the normal vector is }\overrightarrow{n}=\widehat{i}+\widehat{j}-\widehat{k}\text{ and }d=7$
$\displaystyle \text{Now, }\overrightarrow{b}\cdot\overrightarrow{n}=(\widehat{i}+3\widehat{j}+4\widehat{k})\cdot(\widehat{i}+\widehat{j}-\widehat{k})=1+3-4=4-4=0$
$\displaystyle \text{So, }\overrightarrow{b}\text{ is perpendicular to }\overrightarrow{n}$
$\displaystyle \text{So, the given line is parallel to the given plane}$
$\displaystyle \text{The distance between the line and the parallel plane. Then,}$
$\displaystyle d=\text{length of the perpendicular from the point }\overrightarrow{a}=2\widehat{i}+5\widehat{j}+7\widehat{k}\text{ to the plane }\overrightarrow{r}\cdot\overrightarrow{n}=d$
$\displaystyle d=\frac{\left|\overrightarrow{a}\cdot\overrightarrow{n}-d\right|}{\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{\left|(2\widehat{i}+5\widehat{j}+7\widehat{k})\cdot(\widehat{i}+\widehat{j}-\widehat{k})-7\right|}{\left|\widehat{i}+\widehat{j}-\widehat{k}\right|}$
$\displaystyle =\frac{\left|2+5-7-7\right|}{\sqrt{1+1+1}}$
$\displaystyle =\frac{7}{\sqrt{3}}\text{ units}$

$\displaystyle \textbf{Question 6: }~\text{Find the vector equation of the line through the origin which is} \\ \text{perpendicular to the plane }\overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})=3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The required line is perpendicular to the plane }\overrightarrow{r}\cdot(\widehat{i}-2\widehat{j}+3\widehat{k})=3$
$\displaystyle \text{Therefore, it is parallel to the normal }\widehat{i}-2\widehat{j}+3\widehat{k}$
$\displaystyle \text{Thus, the required line passes through the point with position vector }\overrightarrow{a}=0\widehat{i}+0\widehat{j}+0\widehat{k}\text{ and is parallel to the vector }\overrightarrow{n}=\widehat{i}-2\widehat{j}+3\widehat{k}$
$\displaystyle \text{So, its vector equation is}$
$\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{n}$
$\displaystyle \Rightarrow \overrightarrow{r}=0\widehat{i}+0\widehat{j}+0\widehat{k}+\lambda(\widehat{i}-2\widehat{j}+3\widehat{k})$
$\displaystyle \Rightarrow \overrightarrow{r}=\lambda(\widehat{i}-2\widehat{j}+3\widehat{k})$

$\displaystyle \textbf{Question 7: }~\text{Find the equation of the plane through }(2,3,-4)\text{ and } \\ (1,-1,3)\text{ and parallel to }x\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane through }(2,3,-4)\text{ is}$
$\displaystyle a(x-2)+b(y-3)+c(z+4)=0\ldots(1)$
$\displaystyle \text{This plane passes through }(1,-1,3).\text{ So ,}$
$\displaystyle a(1-2)+b(-1-3)+c(3+4)=0$
$\displaystyle \Rightarrow -a-4b+7c=0\ldots(2)$
$\displaystyle \text{Again plane (1) is parallel to the }x\text{-axis. It means that plane (1) is perpendicular to the }yz\text{-plane whose equation is }x=0\text{ or }1\cdot x+0\cdot y+0\cdot z=0$
$\displaystyle \Rightarrow a(1)+b(0)+c(0)=0\ldots(3)\text{ (Because }a_1a_2+b_1b_2+c_1c_2=0\text{)}$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle \left|\begin{array}{ccc}x-3&y-3&z+4\\-1&-4&7\\1&0&0\end{array}\right|=0$
$\displaystyle \Rightarrow 0(x-3)+7(y-3)+4(z+4)=0$
$\displaystyle \Rightarrow 7y+4z-5=0$

$\displaystyle \textbf{Question 8: }~\text{Find the equation of a plane passing through the points }(0,0,0)\text{ and } \\ (3,-1,2)\text{ and parallel to the line }\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}. \text{[CBSE\ 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane through }(0,0,0)\text{ is}$
$\displaystyle a(x-0)+b(y-0)+c(z-0)=0$
$\displaystyle ax+by+cz=0\ldots(1)$
$\displaystyle \text{This plane passes through }(3,-1,2).\text{ So ,}$
$\displaystyle 3a-b+2c=0\ldots(2)$
$\displaystyle \text{Again plane (1) is parallel to the given line}$
$\displaystyle \text{It means that the normal to plane (1) is perpendicular to the line}$
$\displaystyle \Rightarrow a(1)+b(-4)+c(7)=0\ldots(3)\text{ (Because }a_1a_2+b_1b_2+c_1c_2=0\text{)}$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle \left|\begin{array}{ccc}x&y&z\\3&-1&2\\1&-4&7\end{array}\right|=0$
$\displaystyle \Rightarrow x-19y-11z=0$

$\displaystyle \textbf{Question 9: }~\text{Find the vector and cartesian equations of the line passing through }(1,2,3) \\ \text{ and parallel to the planes }\overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+2\widehat{k})=5\text{ and }\overrightarrow{r}\cdot(3\widehat{i}+\widehat{j}+2\widehat{k})=6. \text{[CBSE\ 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the equation of line passing through }(1,2,3)\text{ is given by}$
$\displaystyle \frac{x-1}{a_1}=\frac{y-2}{b_1}=\frac{z-3}{c_1}\ldots(1)$
$\displaystyle \text{We know that line }\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\text{ is parallel to plane }a_2x+b_2y+c_2z+d_2=0\text{ if }a_1a_2+b_1b_2+c_1c_2=0\ldots(2)$
$\displaystyle \text{Here, line (1) is parallel to plane }x-y+2z=5$
$\displaystyle \text{So,}$
$\displaystyle a\times1+b\times(-1)+c\times2=0$
$\displaystyle \Rightarrow a-b+2c=0\ldots(3)$
$\displaystyle \text{Also, line (1) is parallel to plane }3x+y+z=6$
$\displaystyle \text{So,}$
$\displaystyle a\times3+b\times1+c\times1=0$
$\displaystyle \Rightarrow 3a+b+c=0\ldots(4)$
$\displaystyle \text{Solving equations (3) and (4) by cross multiplication we have,}$
$\displaystyle \frac{a}{(-1)(1)-(1)(2)}=\frac{b}{(3)(2)-(1)(1)}=\frac{c}{(1)(1)-(3)(-1)}$
$\displaystyle \Rightarrow \frac{a}{-1-2}=\frac{b}{6-1}=\frac{c}{1+3}$
$\displaystyle \Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{4}=k\text{ (let)}$
$\displaystyle \therefore a=-3k,\;b=5k,\;c=4k$
$\displaystyle \text{Putting the value in equation (1)}$
$\displaystyle \frac{x-1}{-3k}=\frac{y-2}{5k}=\frac{z-3}{4k}$
$\displaystyle \text{Multiplying by }k\text{ we have}$
$\displaystyle \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
$\displaystyle \text{The required equation is}$
$\displaystyle \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
$\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(-3\widehat{i}+5\widehat{j}+4\widehat{k})$

$\displaystyle \textbf{Question 10: }\text{Prove that the line of section of planes }5x+2y-4z+2=0\\ \text{ and }2x+8y+2z-1=0\text{ is parallel to the plane }4x-2y-5z-2=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a,b,c\text{ be the direction ratios of the line of section of the given planes}$
$\displaystyle \text{As this line lies on both the planes, their normals are perpendicular to it}$
$\displaystyle \Rightarrow 5a+2b-4c=0\ldots(1)$
$\displaystyle 2a+8b+2c=0$
$\displaystyle \Rightarrow a+4b+c=0\ldots(2)$
$\displaystyle \text{Using cross-multiplication method, we get}$
$\displaystyle \frac{a}{2+16}=\frac{b}{-4-5}=\frac{c}{20-2}$
$\displaystyle \Rightarrow \frac{a}{18}=\frac{b}{-9}=\frac{c}{18}$
$\displaystyle \Rightarrow \frac{a}{2}=\frac{b}{-1}=\frac{c}{2}$
$\displaystyle \text{So, the direction ratios of the line are proportional to }2,-1,2$
$\displaystyle \text{Direction ratios of the given line are }4,-2,-5$
$\displaystyle \text{Now,}$
$\displaystyle (2)(4)+(-1)(-2)+(2)(-5)$
$\displaystyle =8+2-10$
$\displaystyle =0$
$\displaystyle \text{So, the line of section of the given planes is parallel to the given plane}$

$\displaystyle \textbf{Question 11: }~\text{Find the vector equation of the line passing through }(1,-1,2) \\ \text{ and perpendicular to the plane }2x-y+3z-5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a,b,c\text{ be the direction ratios of the given line}$
$\displaystyle \text{Since the line passes through the point }(1,-1,2)\text{ is}$
$\displaystyle \frac{x-1}{a}=\frac{y+1}{b}=\frac{z-2}{c}\ldots(1)$
$\displaystyle \text{Since this line is perpendicular to the plane }2x-y+3z-5=0,\text{ the line is parallel to the normal of the plane}$
$\displaystyle \text{So, the direction ratios of the line are proportional to the direction ratios of the given plane}$
$\displaystyle \text{So, }\frac{a}{2}=\frac{b}{-1}=\frac{c}{3}=\lambda$
$\displaystyle \Rightarrow a=2\lambda,\;b=-\lambda,\;c=3\lambda$
$\displaystyle \text{Substituting these values in (1), we get}$
$\displaystyle \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{3},\text{ which is the Cartesian form of the line}$
$\displaystyle \text{Vector form}$
$\displaystyle \text{The given line passes through a point whose position vector is }\overrightarrow{a}=\widehat{i}-\widehat{j}+2\widehat{k}\text{ and is parallel to the vector }\overrightarrow{b}=2\widehat{i}-\widehat{j}+3\widehat{k}$
$\displaystyle \text{So, its equation in vector form is}$
$\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
$\displaystyle \Rightarrow \overrightarrow{r}=(\widehat{i}-\widehat{j}+2\widehat{k})+\lambda(2\widehat{i}-\widehat{j}+3\widehat{k})$

$\displaystyle \textbf{Question 12: }~\text{Find the equation of the plane through the points }(2,2,-1)\text{ and } \\ (3,4,2)\text{ and parallel to the line whose direction ratios are }7,0,6.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane through }(2,2,-1)\text{ is}$
$\displaystyle a(x-2)+b(y-2)+c(z+1)=0\ldots(1)$
$\displaystyle \text{This plane passes through }(3,4,2).\text{ So ,}$
$\displaystyle a(3-2)+b(4-2)+c(2+1)=0$
$\displaystyle \Rightarrow a+2b+3c=0\ldots(2)$
$\displaystyle \text{Again plane (1) is parallel to the line whose direction ratios are }7,0,6$
$\displaystyle \text{It means that the normal of plane (1) is perpendicular to the line whose direction ratios are }7,0,6$
$\displaystyle \Rightarrow 7a+0b+6c=0\ldots(3)\text{ (Because }a_1a_2+b_1b_2+c_1c_2=0\text{)}$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle \left|\begin{array}{ccc}x-2&y-2&z+1\\1&2&3\\7&0&6\end{array}\right|=0$
$\displaystyle \Rightarrow 12(x-2)+15(y-2)-14(z+1)=0$
$\displaystyle \Rightarrow 12x+15y-14z-68=0$

$\displaystyle \textbf{Question 13: }~\text{Find the angle between the line }\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2} \\ \text{ and the plane }3x+4y+z+5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given line is parallel to the vector }\overrightarrow{b}=3\widehat{i}-\widehat{j}+2\widehat{k}\text{ and the given plane is normal to the vector }\overrightarrow{n}=3\widehat{i}+4\widehat{j}+\widehat{k}$
$\displaystyle \text{We know that the angle }\theta\text{ between the line and the plane is given by}$
$\displaystyle \sin\theta=\frac{\overrightarrow{b}\cdot\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{(3\widehat{i}-\widehat{j}+2\widehat{k})\cdot(3\widehat{i}+4\widehat{j}+\widehat{k})}{\left|3\widehat{i}-\widehat{j}+2\widehat{k}\right|\left|3\widehat{i}+4\widehat{j}+\widehat{k}\right|}$
$\displaystyle =\frac{9-4+2}{\sqrt{9+1+4}\sqrt{9+16+1}}$
$\displaystyle =\frac{7}{\sqrt{14}\sqrt{26}}$
$\displaystyle =\frac{7}{\sqrt{364}}$
$\displaystyle =\sqrt{\frac{7}{52}}$
$\displaystyle \Rightarrow \theta=\sin^{-1}\!\left(\sqrt{\frac{7}{52}}\right)$

$\displaystyle \textbf{Question 14: }~\text{Find the equation of the plane passing through the intersection of the planes } \\ x-2y+z=1\text{ and }2x+y+z=8\text{ and parallel to the line with direction ratios proportional to } \\ 1,2,1.\text{ Find also the perpendicular distance of }(1,1,1)\text{ from this plane. \ [CBSE\ 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane passing through the intersection of the given planes is}$
$\displaystyle (x-2y+z-1)+\lambda(2x+y+z-8)=0$
$\displaystyle \Rightarrow (1+2\lambda)x+(-2+\lambda)y+(1+\lambda)z-1-8\lambda=0\ldots(1)$
$\displaystyle \text{This plane is parallel to the line whose direction ratios are proportional to }1,2,1$
$\displaystyle \text{So, the normal to the plane is perpendicular to the line whose direction ratios are proportional to }1,2,1$
$\displaystyle \Rightarrow (1+2\lambda)(1)+(-2+\lambda)(2)+(1+\lambda)(1)=0$
$\displaystyle \Rightarrow 1+2\lambda-4+2\lambda+1+\lambda=0$
$\displaystyle \Rightarrow 5\lambda-2=0$
$\displaystyle \Rightarrow \lambda=\frac{2}{5}$
$\displaystyle \text{Substituting this in (1), we get}$
$\displaystyle \left(1+2\left(\frac{2}{5}\right)\right)x+\left(-2+\left(\frac{2}{5}\right)\right)y+\left(1+\left(\frac{2}{5}\right)\right)z-1-8\left(\frac{2}{5}\right)=0$
$\displaystyle \Rightarrow 9x-8y+7z-21=0\ldots(2),\text{ which is the required equation of the plane}$
$\displaystyle \text{Perpendicular distance of plane (2) from }(1,1,1)$
$\displaystyle =\frac{\left|9(1)-8(1)+7(1)-21\right|}{\sqrt{9^{2}+(-8)^{2}+7^{2}}}$
$\displaystyle =\frac{\left|-13\right|}{\sqrt{194}}$
$\displaystyle =\frac{13}{\sqrt{194}}\text{ units}$

$\displaystyle \textbf{Question 15: }~\text{State when the line }\overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}\text{ is parallel to the plane }\overrightarrow{r}\cdot\overrightarrow{n}=d.\text{ Show that the line }\overrightarrow{r}=\widehat{i}+\lambda(3\widehat{i}-\widehat{j}+2\widehat{k})\text{ is parallel to the plane }\overrightarrow{r}\cdot(2\widehat{i}+\widehat{k})=3.\text{ Also, find the distance between the line and the plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given plane passes through the point with position vector }\overrightarrow{a}=\widehat{i}+\widehat{j}+0\widehat{k}\text{ and is parallel to the vector }\overrightarrow{b}=3\widehat{i}-\widehat{j}+2\widehat{k}$
$\displaystyle \text{The given plane is }\overrightarrow{r}\cdot(2\widehat{j}+\widehat{k})=3\text{ or }\overrightarrow{r}\cdot\overrightarrow{n}=d$
$\displaystyle \text{So, normal vector }\overrightarrow{n}=0\widehat{i}+2\widehat{j}+\widehat{k}\text{ and }d=3$
$\displaystyle \text{Now, }\overrightarrow{b}\cdot\overrightarrow{n}=(3\widehat{i}-\widehat{j}+2\widehat{k})\cdot(0\widehat{i}+2\widehat{j}+\widehat{k})=0-2+2=0$
$\displaystyle \text{So, }\overrightarrow{b}\text{ is perpendicular to }\overrightarrow{n}$
$\displaystyle \text{Hence, the given line is parallel to the given plane}$
$\displaystyle \text{The distance between the line and the parallel plane is the distance between} \\ \text{any point on the line and the given plane. The plane passes through the point } \\ \overrightarrow{a}=\widehat{i}+\widehat{j}+0\widehat{k}$
$\displaystyle \text{The perpendicular distance from the given line to the plane is}$
$\displaystyle d=\frac{\left|\overrightarrow{a}\cdot\overrightarrow{n}-d\right|}{\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{\left|(\widehat{i}+\widehat{j}+0\widehat{k})\cdot(0\widehat{i}+2\widehat{j}+\widehat{k})-3\right|}{\left|0\widehat{i}+2\widehat{j}+\widehat{k}\right|}$
$\displaystyle =\frac{\left|0+2+0-3\right|}{\sqrt{0^{2}+2^{2}+1^{2}}}$
$\displaystyle =\frac{1}{\sqrt{5}}\text{ units}$

$\displaystyle \textbf{Question 16: }~\text{Show that the plane whose vector equation is }\overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}-\widehat{k})=1\text{ and the line whose vector equation is }\overrightarrow{r}=(-\widehat{i}+\widehat{j}+\widehat{k})+\lambda(2\widehat{i}+\widehat{j}+4\widehat{k})\text{ are parallel. Also, find the distance between them.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given plane passes through the point with position vector }\overrightarrow{a}=-\widehat{i}+\widehat{j}+\widehat{k}\text{ and is parallel to the vector }\overrightarrow{b}=2\widehat{i}+\widehat{j}+4\widehat{k}$
$\displaystyle \text{The given plane is }\overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}-\widehat{k})=1\text{ or }\overrightarrow{r}\cdot\overrightarrow{n}=d$
$\displaystyle \text{So, normal vector }\overrightarrow{n}=\widehat{i}+2\widehat{j}-\widehat{k}\text{ and }d=1$
$\displaystyle \text{Now, }\overrightarrow{b}\cdot\overrightarrow{n}=(2\widehat{i}+\widehat{j}+4\widehat{k})\cdot(\widehat{i}+2\widehat{j}-\widehat{k})=2+2-4=0$
$\displaystyle \text{So, }\overrightarrow{b}\text{ is perpendicular to }\overrightarrow{n}$
$\displaystyle \text{So, the given line is parallel to the given plane}$
$\displaystyle \text{The distance between the line and the parallel plane is the distance between} \\ \text{any point on the line and the given plane. Since the plane passes through the point } \\ \overrightarrow{a}=-\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{the perpendicular distance from the given line to the plane is}$
$\displaystyle d=\frac{\left|\overrightarrow{a}\cdot\overrightarrow{n}-d\right|}{\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{\left|(-\widehat{i}+\widehat{j}+\widehat{k})\cdot(\widehat{i}+2\widehat{j}-\widehat{k})-1\right|}{\left|\widehat{i}+2\widehat{j}-\widehat{k}\right|}$
$\displaystyle =\frac{\left|-1+2-1-1\right|}{\sqrt{1+4+1}}$
$\displaystyle =\frac{1}{\sqrt{6}}\text{ units}$

$\displaystyle \textbf{Question 17: }~\text{Find the equation of the plane through the intersection} \\ \text{of the planes }3x-4y+5z=10\text{ and }2x+2y-3z=4\text{ and parallel to the line }x=2y=3z.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane passing through the intersection of the given planes is}$
$\displaystyle (3x-4y+5z-10)+\lambda(2x+2y-3z-4)=0$
$\displaystyle \Rightarrow (3+2\lambda)x+(-4+2\lambda)y+(5-3\lambda)z-10-4\lambda=0\ldots(1)$
$\displaystyle \text{The given line is}$
$\displaystyle x=2y=3z$
$\displaystyle \text{Dividing this equation by }6,\text{ we get}$
$\displaystyle \frac{x}{6}=\frac{y}{3}=\frac{z}{2}$
$\displaystyle \text{The direction ratios of this line are proportional to }6,3,2$
$\displaystyle \text{So, the normal to the plane is perpendicular to the line whose direction ratios} \\ \text{are proportional to }6,3,2$
$\displaystyle \Rightarrow (3+2\lambda)6+(-4+2\lambda)3+(5-3\lambda)2=0$
$\displaystyle \Rightarrow 18+12\lambda-12+6\lambda+10-6\lambda=0$
$\displaystyle \Rightarrow 12\lambda+16=0$
$\displaystyle \Rightarrow \lambda=-\frac{4}{3}$
$\displaystyle \text{Substituting this in (1), we get}$
$\displaystyle \left(3+2\left(-\frac{4}{3}\right)\right)x+\left(-4+2\left(-\frac{4}{3}\right)\right)y+\left(5-3\left(-\frac{4}{3}\right)\right)z-10-4\left(-\frac{4}{3}\right)=0$
$\displaystyle \Rightarrow x-20y+27z=14$

$\displaystyle \textbf{Question 18: }~\text{Find the vector and cartesian forms of the equation of the} \\ \text{plane passing through } (1,2,-4)\text{ and parallel to the lines }\overrightarrow{r}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k})\text{ and } \overrightarrow{r}=(\widehat{i}-3\widehat{j}+5\widehat{k})+\mu(\widehat{i}-\widehat{k}).\text{ Also, find the distance of the point } \\ (9,-8,-10) \text{ from the plane thus obtained. \ [CBSE\ 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given lines are }\overrightarrow{r}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k}),\;\overrightarrow{r}=(\widehat{i}-3\widehat{j}+5\widehat{k})+\mu(\widehat{i}+\widehat{j}-\widehat{k})$
$\displaystyle \text{We know that the vector equation of a plane passing through a point }\overrightarrow{a} \\ \text{ and parallel to }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ is given by }(\overrightarrow{r}-\overrightarrow{a})\cdot(\overrightarrow{b}\times\overrightarrow{c})=0$
$\displaystyle \text{Here, }\overrightarrow{a}=\widehat{i}+2\widehat{j}-4\widehat{k}$
$\displaystyle \overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}\text{ and}$
$\displaystyle \overrightarrow{c}=\widehat{i}+\widehat{j}-\widehat{k}$
$\displaystyle \therefore \overrightarrow{b}\times\overrightarrow{c}=\left|\begin{array}{ccc}\widehat{i}&\widehat{j}&\widehat{k}\\2&3&6\\1&1&-1\end{array}\right|=-9\widehat{i}+8\widehat{j}-\widehat{k}$
$\displaystyle \text{So, the vector equation of the plane is }(\overrightarrow{r}-\overrightarrow{a})\cdot(\overrightarrow{b}\times\overrightarrow{c})=0$
$\displaystyle \Rightarrow [\overrightarrow{r}-(\widehat{i}+2\widehat{j}-4\widehat{k})]\cdot(-9\widehat{i}+8\widehat{j}-\widehat{k})=0$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-9\widehat{i}+8\widehat{j}-\widehat{k})=(\widehat{i}+2\widehat{j}-4\widehat{k})\cdot(-9\widehat{i}+8\widehat{j}-\widehat{k})$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-9\widehat{i}+8\widehat{j}-\widehat{k})=1(-9)+2(8)+(-4)(-1)=-9+16+4=11$
$\displaystyle \text{Thus, the vector equation of the plane is }\overrightarrow{r}\cdot(-9\widehat{i}+8\widehat{j}-\widehat{k})=11$
$\displaystyle \text{The Cartesian equation of this plane is }(x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(-9\widehat{i}+8\widehat{j}-\widehat{k})=11$
$\displaystyle \Rightarrow -9x+8y-z=11$
$\displaystyle \text{Now,}$
$\displaystyle \text{Distance of the point }(9,-8,-10)\text{ from the plane }-9x+8y-z=11$
$\displaystyle =\text{Length of perpendicular from }(9,-8,-10)\text{ from the plane }-9x+8y-z-11=0$
$\displaystyle =\frac{\left|-9(9)+8(-8)-(-10)-11\right|}{\sqrt{(-9)^{2}+8^{2}+(-1)^{2}}}$
$\displaystyle =\frac{\left|-81-64+10-11\right|}{\sqrt{81+64+1}}$
$\displaystyle =\frac{\left|-146\right|}{\sqrt{146}}$
$\displaystyle =\sqrt{146}\text{ units}$

$\displaystyle \textbf{Question 19: }~\text{Find the equation of the plane passing through the points }(3,4,1) \\ \text{ and }(0,1,0)\text{ and parallel to the line }\frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}. \text{[CBSE\ 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane passing through }(3,4,1)\text{ is}$
$\displaystyle a(x-3)+b(y-4)+c(z-1)=0\ldots(1)$
$\displaystyle \text{This plane passes through }(0,1,0).\text{ So ,}$
$\displaystyle a(0-3)+b(1-4)+c(0-1)=0$
$\displaystyle \Rightarrow -3a-3b-c=0$
$\displaystyle \Rightarrow 3a+3b+c=0\ldots(2)$
$\displaystyle \text{Again plane (1) is parallel to the given line}$
$\displaystyle \text{It means that the normal to plane (1) is perpendicular to the line}$
$\displaystyle \Rightarrow a(2)+b(7)+c(5)=0\ldots(3)\text{ (Because }a_1a_2+b_1b_2+c_1c_2=0\text{)}$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle \left|\begin{array}{ccc}x-3&y-4&z-1\\3&3&1\\2&7&5\end{array}\right|=0$
$\displaystyle \Rightarrow 8(x-3)-13(y-4)+15(z-1)=0$
$\displaystyle \Rightarrow 8x-13y+15z+13=0$

$\displaystyle \textbf{Question 20: }~\text{Find the coordinates of the point where the line } \\ \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2} \text{ intersects the plane }x-y+z-5=0. \\ \text{ Also, find the angle between the line and the plane. \ [CBSE\ 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda\text{ (say)}$
$\displaystyle \Rightarrow x=3\lambda+2;\;y=4\lambda-1;\;z=2\lambda+2\ldots(1)$
$\displaystyle \text{Since }(x,y,z)\text{ intersects the plane }x-y+z-5=0,$
$\displaystyle 3\lambda+2-(4\lambda-1)+2\lambda+2-5=0$
$\displaystyle \Rightarrow 3\lambda+2-4\lambda+1+2\lambda+2-5=0$
$\displaystyle \Rightarrow \lambda=0$
$\displaystyle \text{Substituting this in (1), we get}$
$\displaystyle x=2;\;y=-1;\;z=2$
$\displaystyle \text{So, }(x,y,z)=(2,-1,2)$
$\displaystyle \text{Finding the angle}$
$\displaystyle \text{The given line is parallel to the vector }\overrightarrow{b}=3\widehat{i}+4\widehat{j}+2\widehat{k} \\ \text{ and the given plane is normal to the vector }\overrightarrow{n}=\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \text{We know that the angle }\theta\text{ between a line and a plane is given by}$
$\displaystyle \sin\theta=\frac{\overrightarrow{b}\cdot\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{(3\widehat{i}+4\widehat{j}+2\widehat{k})\cdot(\widehat{i}-\widehat{j}+\widehat{k})}{\left|3\widehat{i}+4\widehat{j}+2\widehat{k}\right|\left|\widehat{i}-\widehat{j}+\widehat{k}\right|}$
$\displaystyle =\frac{3-4+2}{\sqrt{9+16+4}\sqrt{1+1+1}}$
$\displaystyle =\frac{1}{\sqrt{87}}$
$\displaystyle \Rightarrow \theta=\sin^{-1}\!\left(\frac{1}{\sqrt{87}}\right)$

$\displaystyle \textbf{Question 21: }~\text{Find the vector equation of the line passing through }(1,2,3) \\ \text{ and perpendicular to the plane }\overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a,b,c\text{ be the direction ratios of the given line}$
$\displaystyle \text{Since the line passes through the point }(1,2,3)\text{ is ,}$
$\displaystyle \frac{x-1}{a}=\frac{y-2}{b}=\frac{z-3}{c}\ldots(1)$
$\displaystyle \text{Since this line is perpendicular to the plane }\overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0\text{ or }x+2y-5z+9=0,\text{ the line is parallel to the normal of the plane}$
$\displaystyle \text{So, the direction ratios of the line are proportional to the direction ratios of the given plane}$
$\displaystyle \text{So, }\frac{a}{1}=\frac{b}{2}=\frac{c}{-5}=\lambda$
$\displaystyle \Rightarrow a=\lambda,\;b=2\lambda,\;c=-5\lambda$
$\displaystyle \text{Substituting these values in (1), we get}$
$\displaystyle \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5},\text{ which is the Cartesian form of the line}$
$\displaystyle \text{Vector form}$
$\displaystyle \text{The given line passes through a point whose position vector is }\overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k}\text{ and is parallel to the vector }\overrightarrow{b}=\widehat{i}+2\widehat{j}-5\widehat{k}$
$\displaystyle \text{So, its equation in vector form is}$
$\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
$\displaystyle \Rightarrow \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(\widehat{i}+2\widehat{j}-5\widehat{k})$

$\displaystyle \textbf{Question 22: }~\text{Find the angle between the line }\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6} \\ \text{ and the plane }10x+2y-11z=3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given line is parallel to the vector }\overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k} \\ \text{ and the given plane is normal to the vector }\overrightarrow{n}=10\widehat{i}+2\widehat{j}-11\widehat{k}$
$\displaystyle \text{We know that the angle }\theta\text{ between the line and the plane is given by}$
$\displaystyle \sin\theta=\frac{\overrightarrow{b}\cdot\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{(2\widehat{i}+3\widehat{j}+6\widehat{k})\cdot(10\widehat{i}+2\widehat{j}-11\widehat{k})}{\left|2\widehat{i}+3\widehat{j}+6\widehat{k}\right|\left|10\widehat{i}+2\widehat{j}-11\widehat{k}\right|}$
$\displaystyle =\frac{20+6-66}{\sqrt{4+9+36}\sqrt{100+4+121}}$
$\displaystyle =\frac{-40}{7\cdot15}$
$\displaystyle =-\frac{8}{21}$
$\displaystyle \Rightarrow \theta=\sin^{-1}\!\left(-\frac{8}{21}\right)$

$\displaystyle \textbf{Question 23: }~\text{Find the vector equation of the line passing through }(1,2,3) \\\text{ and parallel to the planes }\overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+2\widehat{k})=5 \\ \text{ and }\overrightarrow{r}\cdot(3\widehat{i}+\widehat{j}+\widehat{k})=6.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given line passes through }(1,2,3)\Rightarrow \overrightarrow a=\widehat i+2\widehat j+3\widehat k$
$\displaystyle \text{Given planes: }\overrightarrow r\cdot(\widehat i-\widehat j+2\widehat k)=5\text{ and }\overrightarrow r\cdot(3\widehat i+\widehat j+\widehat k)=6$
$\displaystyle \text{Normal vectors of the planes are }\overrightarrow n_1=(1,-1,2)\text{ and }\overrightarrow n_2=(3,1,1)$
$\displaystyle \text{A line parallel to both planes must have direction vector perpendicular to both }\overrightarrow n_1\text{ and }\overrightarrow n_2$
$\displaystyle \Rightarrow \text{direction vector } \overrightarrow b=\overrightarrow n_1\times \overrightarrow n_2$
$\displaystyle \overrightarrow b=\begin{vmatrix}\widehat i&\widehat j&\widehat k\\1&-1&2\\3&1&1\end{vmatrix}$
$\displaystyle =\widehat i((-1)(1)-2(1))-\widehat j(1(1)-2(3))+\widehat k(1(1)-(-1)(3))$
$\displaystyle =\widehat i(-1-2)-\widehat j(1-6)+\widehat k(1+3)$
$\displaystyle =-3\widehat i+5\widehat j+4\widehat k$
$\displaystyle \text{Hence vector equation of the line is }\overrightarrow r=\overrightarrow a+\lambda\overrightarrow b$
$\displaystyle \therefore\ \overrightarrow r=(\widehat i+2\widehat j+3\widehat k)+ \lambda(-3\widehat i+5\widehat j+4\widehat k)$

$\displaystyle \textbf{Question 24: }~\text{Find the value of }\lambda\text{ such that the line }\frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{-4} \\ \text{ is perpendicular to the plane }3x-y-2z=7. \text{[CBSE\ 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given line }\frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{-4}$
$\displaystyle \text{Direction ratios of the line }=(6,\lambda,-4)$
$\displaystyle \text{Given plane }3x-y-2z=7$
$\displaystyle \text{Normal vector of the plane }=(3,-1,-2)$
$\displaystyle \text{For the line to be perpendicular to the plane, its direction ratios must be parallel to the plane's normal}$
$\displaystyle \Rightarrow (6,\lambda,-4)=k(3,-1,-2)$
$\displaystyle \text{So }\frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}$
$\displaystyle 2=-\lambda=2$
$\displaystyle \lambda=-2$

$\displaystyle \textbf{Question 25: }~\text{Find the equation of the plane passing through the points }(-1,2,0), \\ (2,2,-1) \text{ and parallel to the line }\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}. \text{[CBSE\ 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The general equation of the plane passing through the point }(-1,2,0) \\ \text{ is given by }a(x+1)+b(y-2)+c(z-0)=0\ldots(1)$
$\displaystyle \text{If this plane passes through the point }(2,2,-1),\text{ we have}$
$\displaystyle a(2+1)+b(2-2)+c(-1-0)=0$
$\displaystyle \Rightarrow 3a-c=0\ldots(2)$
$\displaystyle \text{Direction ratios of the normal to the plane (1) are }a,b,c$
$\displaystyle \text{The equation of the given line is}$
$\displaystyle \frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$
$\displaystyle \text{This can be rewritten as}$
$\displaystyle \frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$
$\displaystyle \text{Direction ratios of the line are }1,1,-1$
$\displaystyle \text{The required plane is parallel to the given line when the normal to this} \\ \text{plane is perpendicular to this line}$
$\displaystyle \Rightarrow a(1)+b(1)+c(-1)=0$
$\displaystyle \Rightarrow a+b-c=0\ldots(3)$
$\displaystyle \text{Solving (2) and (3), we get}$
$\displaystyle \frac{a}{0+1}=\frac{b}{-1+3}=\frac{c}{3-0}$
$\displaystyle \Rightarrow \frac{a}{1}=\frac{b}{2}=\frac{c}{3}=\lambda\text{ (say)}$
$\displaystyle \Rightarrow a=\lambda,\;b=2\lambda,\;c=3\lambda$
$\displaystyle \text{Putting these values of }a,b,c\text{ in (1), we have}$
$\displaystyle \lambda(x+1)+2\lambda(y-2)+3\lambda(z-0)=0$
$\displaystyle \Rightarrow x+1+2y-4+3z=0$
$\displaystyle \Rightarrow x+2y+3z=3$
$\displaystyle \text{Thus, the equation of the required plane is }x+2y+3z=3$