Exercise 6.9-Additional Problems - Chapter 6 - Applications of Vector Algebra - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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​Additional Problems
Question 1.
Find the point of intersection of the line passing through the two points $(1,1,-1) ;(-1,0,1)$ and the xy-plane.

Solution:
The equation of the line passing through $(1,1,-1)$ and $(-1,0,1)$ is
$
\frac{x-1}{2}=\frac{y-1}{1}=\frac{z+1}{-2}
$
It meets the $x y$-plane i.e. $z=0$
$
\therefore \quad \frac{x-1}{2}=\frac{y-1}{1}=\frac{1}{-2} \quad \Rightarrow x=0, y=\frac{1}{2}
$
The required point is $\left(0, \frac{1}{2}, 0\right)$.

Question 2.
Find the co-ordinates of the point where the line $\vec{r}=(\vec{i}+2 \vec{j}-5 \vec{k})+t(2 \vec{i}-3 \vec{j}+4 \vec{k})$ meets the plane $\vec{r} \cdot(2 \vec{i}+4 \vec{j}-\vec{k})=3$.
Solution:
The equation of the straight line in the cartesian form is
$\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}=\lambda$ (say)
$\therefore$ Any point on this line is of the form $(2 \lambda+1,-3 \lambda,+2,4 \lambda-5)$
The cartesian equation of the plane is $2 \mathrm{x}+4 \mathrm{y}-\mathrm{z}-3=0$
But the required point lies on this plane.
$\therefore 2(2 \lambda+1)+4(-3 \lambda+2)-(4 \lambda-5)-3=0 \Rightarrow \lambda=1$
$\therefore$ The required point is $(3,-1,-1)$.
Question 3.
Find the point of intersection of the line $\vec{r}=(\vec{j}-\vec{k})+s(2 \vec{i}-\vec{j}+\vec{k})$ and xz-plane.
Solution:
The given point is $(0,1,-1)$; parallel vector is $2 \vec{i}-\vec{j}+\vec{k}$.
The equation of the line passing through the point $(0,1,-1)$ and parallel to the vector $2 \vec{i}-\vec{j}+\vec{k}$ is
$
\frac{x-0}{2}=\frac{y-1}{-1}=\frac{z+1}{1}
$
It meets the $x z$-plane i.e. $y=0$
$
\Rightarrow \begin{array}{rlr}
\frac{x}{2}=1=\frac{z+1}{1} & \Rightarrow \frac{x}{2}=1 \Rightarrow x=2 \\
\frac{z+1}{1}=1 & \Rightarrow z+1=1 \Rightarrow z=0
\end{array}
$
$\therefore$ The required point is $(2,0,0)$
Question 4.
Find the meeting point of the line $\vec{r}=(2 \vec{i}+\vec{j}-3 \vec{k})+t(2 \vec{i}-\vec{j}-\vec{k})$ and the plane $\mathrm{x}-2 \mathrm{y}+3 \mathrm{z}+7=0$.

Solution:
The equation of the straight line in cartesian form is $\frac{x-2}{2}=\frac{y-1}{-1}=\frac{z+3}{-1}=\lambda$ (say)
Any point on the line is $(2 \lambda,+2,-\lambda+1, \lambda-3)$
The point lies on the plane $x-2 y+3 z+7=0$

$
\begin{aligned}
& \Rightarrow 2 \lambda,+2-2(-\lambda+1)+3(-\lambda,-3)+7=0 . \\
& 2 \lambda+2+2 \lambda-2-3 \lambda-9+7=0 \\
& \lambda-2=0 \Rightarrow \lambda=2 \\
& \text { So } 2 \lambda+2=6 ;-\lambda+1=-1 ;-\lambda-3=-5 \\
& \therefore \text { The required point is }(6,-1,-5)
\end{aligned}
$
Question 5.
Show that the following planes are at right angles:
$
\vec{r} \cdot(2 \vec{i}-\vec{j}+\vec{k})=15 \text { and } \vec{r} \cdot(\vec{i}-\vec{j}-3 \vec{k})=3
$
Solution:
The normal vectors are
$\bar{n}_1=2 \bar{i}-\bar{j}+\bar{k}$ and $\vec{n}_2=+\vec{i}-\vec{j}-3 \vec{k}$
$\vec{n}_1 \cdot \vec{n}_2=(2)(1)+(-1)(-1)+(1)(-3)=2+1-3=0$
$\Rightarrow \vec{n}_1 \perp \mathrm{r}$ to $\vec{n}_2$, i.e., the normals to the planes are at right angles. So, the planes are at right angles.
Question 6.
The planes $\vec{r} \cdot(2 \vec{i}+\lambda \vec{j}-3 \vec{k})=10$ and $\vec{r} \cdot(\lambda \vec{i}+3 \vec{j}+\vec{k})=5$ are perpendicular. Find $\lambda$.
Solution:
Since the planes are perpendicular, the angle between the normals $=90^{\circ}$.
The normals are $\vec{n}_1=2 \vec{i}+\lambda \vec{j}-3 \vec{k}$ and $\vec{n}_2=\lambda \vec{i}+3 \vec{j}+\vec{k}$ $\vec{n}_1 \cdot \vec{n}_2=0[\because \theta=\pi / 2]$
$\Rightarrow(2)(\lambda)+(\lambda)(3)+(-3)(1)=0 \Rightarrow 2 \lambda+3 \lambda-3=0$
$5 \lambda-3=0 \Rightarrow 5 \lambda=3 \Rightarrow \lambda=3 / 5$

Question 7.
Find the angle between the line $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{-2}$ and the plane $3 x+4 y+z=0$.
Solution:
The angle between the line $\vec{r}=\vec{a}+t \vec{b}$ and the plane $\vec{r} \cdot \vec{n}=\mathrm{p}$ is given by the formula
$
\begin{aligned}
\sin \theta & =\frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|} \\
\text { So, } \sin \theta & =\frac{(3 \vec{i}-\vec{j}-2 \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+\vec{k})}{|3 \vec{i}-\vec{j}-2 \vec{k}||3 \vec{i}+4 \vec{j}+\vec{k}|}=\frac{9-4-2}{\sqrt{9+1+4} \sqrt{9+16+1}} \\
& =\frac{3}{\sqrt{14} \sqrt{26}}=\frac{3}{\sqrt{364}}=\frac{3}{\sqrt{4 \times 91}}=\frac{3}{2 \sqrt{91}}
\end{aligned}
$
So, $\theta=\sin ^{-1} \frac{3}{2 \sqrt{91}}$
Question 8.
Find the angle between the line $\vec{r}=\vec{i}+\vec{j}+3 \vec{k}+\lambda(2 \vec{i}+\vec{j}-\vec{k})$ and the plane $\vec{r} \cdot(\vec{i}+\vec{j})=1$,

Solution:
Here $\vec{b}=2 \vec{i}+\vec{j}-\vec{k}$ and $\vec{n}=\vec{i}+\vec{j}$
The angle between the line and the plane is given by
$
\begin{aligned}
& \sin \theta=\frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|} \\
& \vec{b} \cdot \vec{n}=(2 \vec{i}+\vec{j}-\vec{k}) \cdot(\vec{i}+\vec{j})=2+1=3 \\
& \vec{b}=\sqrt{4+1+1}=\sqrt{6} \\
& \vec{n}=\sqrt{1+1}=\sqrt{2} \\
& \therefore \quad \sin \theta=\frac{3}{\sqrt{6} \sqrt{2}}=\frac{3}{\sqrt{12}}=\frac{3}{\sqrt{4 \times 3}}=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{. \operatorname{So} \theta}=\frac{\pi}{3} \\
&
\end{aligned}
$