Exercise 6.9 - Chapter 6 Applications of Vector Algebra 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $6.9$
Question 1.
Find the equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})=3$ and $3 x-5 y+4 z+11=0$, and the point $(-2,1,3)$.
Solution:
Given planes are
$\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})=3$
$2 x-7 y+4 z-3=0$ and $3 x-5 y+4 z+11=0$
Equation of a plane which passes through the line of intersection of the planes
$(2 x-7 y+4 z-3)+\lambda(3 x-5 y+4 z+11)=0$
This passes through the point $(-2,1,3)$.
$
\begin{aligned}
& \text { (1) } \Rightarrow(-4-7+12-3)+\lambda(-6-5+12+11)=0 \\
& -2+\lambda(12)=0 \Rightarrow 12 \lambda=2 \\
& \lambda=\frac{2}{12} \Rightarrow \lambda=\frac{1}{6}
\end{aligned}
$
The required equation is
$
\begin{aligned}
& (1) \Rightarrow(2 x-7 y+4 z-3)+\frac{1}{6}(3 x-5 y+4 z+11)=0 \\
& 12 x-42 y+24 z-18+3 x-5 y+4 z+11=0 \\
& 15 x-47 y+28 z-7=0
\end{aligned}
$
Question 2.
Find the equation of the plane passing through the line of intersection of the planes $x+2 y+3 z=2$ and $x-y$ $+z+11=3$, and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$.
Solution:
Equation of a plane which passes through the line of intersection of the plane
$
(x+2 y+3 z-2)+\lambda(x-y+z-3)=0
$

Question 3.
Find the angle between the line $\vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+t(\hat{i}+2 \hat{j}-2 \hat{k})$ and the plane $\vec{r} \cdot(6 \hat{i}+3 \hat{j}+2 \hat{k})=8$.

Solution:
Angle between the line and a plane
$
\begin{aligned}
\sin \theta & =\frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|} \\
\vec{b} & =\vec{i}+2 \vec{j}-2 \vec{k} \\
\vec{n} & =6 \vec{i}+3 \vec{j}+2 \vec{k} \\
\vec{b} \cdot \vec{n} & =6+6-4=8 \\
|\vec{b}| & =\sqrt{1+4+4}=\sqrt{9}=3 \\
|\vec{n}| & =\sqrt{36+9+4}=\sqrt{49}=7
\end{aligned}
$
$
(1) \Rightarrow \quad \sin \theta=\frac{8}{(3)(7)} \quad \Rightarrow \theta=\sin ^{-1}\left(\frac{8}{21}\right)
$

Question 4.
Find the angle between the planes $\vec{r} \cdot(\hat{i}+\hat{j}-2 \hat{k})=3$ and $2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=2$.
Solution:
Angle between given two planes
$
\begin{aligned}
\cos \theta & =\frac{\left|\vec{n}_1 \cdot \vec{n}_2\right|}{\left|\vec{n}_1\right|\left|\vec{n}_2\right|} \\
\vec{n}_1 & =\vec{i}+\vec{j}-2 \vec{k}, \vec{n}_2=2 \vec{i}-2 \vec{j}+\vec{k} \\
\vec{n}_1 \cdot \vec{n}_2 & =2-2-2=-2 \\
\left|\vec{n}_1\right| & =\sqrt{1+1+4}=\sqrt{6} \\
\left|\vec{n}_2\right| & =\sqrt{4+4+1}=3 \\
\cos \theta & =\frac{|-2|}{3 \sqrt{6}} \quad \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3 \sqrt{6}}\right)
\end{aligned}
$

Question 5.
Find the equation of the plane which passes through the point $(3,4,-1)$ and is parallel to the plane $2 x-3 y+$ $5 z+7=0$. Also, find the distance between the two planes.
Solution:
The required equation parallel to the plane
$
\begin{aligned}
& 2 x-3 y+5 z+7=0 \\
& 2 x-3 y+5 z+\lambda=0
\end{aligned}
$
This passes through $(3,4,-1)$
(2) $\Rightarrow 2(3)-3(4)+5(-1)+\lambda=0$
$6-12-5+1=0$
$\lambda=11$
(2) $\Rightarrow$ The required equation is $2 x-3 y+5 z+11=0$
$\therefore$ Now, distance between the above parallel lines (1) and (3)
$
\begin{aligned}
& \mathrm{a}=2, \mathrm{~b}=-3, \mathrm{c}=5, \mathrm{~d}_1=7, \mathrm{~d}_2=11 \\
& \text { Distance }=\left|\frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}\right|=\left|\frac{11-7}{\sqrt{(2)^2+(-3)^2+(5)^2}}\right|=\left|\frac{4}{\sqrt{4+9+25}}\right| \\
& =\frac{4}{\sqrt{38}} \\
&
\end{aligned}
$

Question 6.
Find the length of the perpendicular from the point $(1,-2,3)$ to the plane $x-y+z=5$.
Solution:
Perpendicular length from the point $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ to the plane $a x+b y+c z+d=0$ is
$
\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|
$
Given point $(1,-2,3)$ and the plane $x-y+z=5$
$
\therefore \text { Length of the perpendicular }=\left|\frac{1-(-2)+3-5}{\sqrt{(1)^2+(-1)^2+(1)^2}}\right|=\left|\frac{1+2+3-5}{\sqrt{3}}\right|=\frac{1}{\sqrt{3}} \text { units }
$

Question 7.
Find the point of intersection of the line $x-1=\frac{y}{2}=z+1$ with the plane $2 x-y+2 z=2$. Also, find the angl between the line and the plane.
Solution:
Any point on the line $\mathrm{x}-1=\frac{y}{2}=\mathrm{z}+1$ is
$
\begin{aligned}
& \mathrm{x}-1=\frac{y}{2}=\mathrm{z}+1=\lambda, \text { (say) } \\
& (\lambda+1,2 \lambda, \lambda-1)
\end{aligned}
$
This passes through the plane $2 x-y+2 z=2$
$
\begin{aligned}
& 2(\lambda+1)-2 \lambda+2(\lambda-1)=2 \\
& 2 \lambda+2-2 \lambda+2 \lambda-2=2 \\
& \lambda=1
\end{aligned}
$
$\therefore$ The required point of intersection is $(2,2,0)$

Angle between the line and the plane is
$
\begin{aligned}
\sin \theta & =\frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|} \\
\vec{b} & =\vec{i}+2 \vec{j}+\vec{k} \text { and } \vec{n}=2 \vec{i}-\vec{j}+2 \vec{k} \\
\vec{b} \cdot \vec{n} & =2-2+2=2 \\
|\vec{b}| & =\sqrt{1+4+1}=\sqrt{6} \\
|\vec{n}| & =\sqrt{4+1+4}=\sqrt{9}=3 \\
\sin \theta & =\frac{2}{3 \sqrt{6}} \quad \Rightarrow \theta=\sin ^{-1}\left(\frac{2}{3 \sqrt{6}}\right)
\end{aligned}
$

Question 8.
Find the coordinates of the foot of the perpendicular and length of the perpendicular from the point $(4,3,2)$ to the plane $x+2 y+3 z=2$
Solution:
Direction of the normal plane $(1,2,3)$

d.c.s of the PQ is $\frac{x_1-4}{1}=\frac{y_1-3}{2}=\frac{z_1-2}{3}=\mathrm{k}$
$\mathrm{x}_1=\mathrm{k}+4, \mathrm{y}_1=2 \mathrm{k}+3, \mathrm{z}_1=3 \mathrm{k}+2$
This passes through the plane $\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=2$
$\mathrm{k}+4+2(2 \mathrm{k}+3)+3(3 \mathrm{k}+2)=2$
$\mathrm{k}+4+4 \mathrm{k}+6+9 \mathrm{k}+6=2$
$14 \mathrm{k}=2-16 \Rightarrow 14 \mathrm{k}=-14$
$\mathrm{k}=-1$
$\therefore$ The coordinate of the foot of the perpendicular is $(3,1,-1)$
$\therefore$ Length of the perpendicular to the plane is
$
=\left|\frac{4+2(3)+3(2)-2}{\sqrt{(1)^2+(2)^2+(3)^2}}\right|=\left|\frac{4+6+6-2}{\sqrt{1+4+9}}\right|=\frac{14}{\sqrt{14}}=\sqrt{14} \text { units }
$​