Exercise 6.6 - 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 $6.6$
Question 1.
Find a parametric form of vector equation of a plane which is at a distance of 7 units from the origin having $3,-4,5$ as direction ratios of a normal to it.
Solution:
Given $\mathrm{p}=7$
Direction ratios : $3,-4,5$ and Direction cosines $\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{5 \sqrt{2}}\right)$
$
\hat{d}=\frac{\vec{d}}{|\vec{d}|}=\frac{1}{5 \sqrt{2}}(3 \vec{i}-4 \vec{j}+5 \vec{k})
$
$\therefore$ The parametric form of vector equation of the plane is
$
\begin{aligned}
\vec{r} \cdot \hat{d} & =p \\
\vec{r} \cdot\left(\frac{3 \vec{i}-4 \vec{j}+5 \vec{k}}{5 \sqrt{2}}\right) & =7
\end{aligned}
$
Question 2.
Find the direction cosines of the normal to the plane $12 x+3 y-4 z=65$. Also, find the non-parametric form of vector equation of a plane and the length of the perpendicular to the plane from the origin.
Solution:
$
\begin{aligned}
12 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z} & =65 \\
\vec{r} \cdot\left(\frac{12 \vec{i}+3 \vec{j}-4 \vec{k}}{\sqrt{144+9+16}}\right) & =\frac{65}{\sqrt{144+9+16}} \\
\vec{r} \cdot\left(\frac{12 \vec{i}+3 \vec{j}-4 \vec{k}}{13}\right) & =\frac{65}{13} \quad \Rightarrow \vec{r} \cdot\left(\frac{12 \vec{i}+3 \vec{j}-4 \vec{k}}{13}\right)=5
\end{aligned}
$
(i) Direction cosines $\left(\frac{12}{13}, \frac{3}{13}, \frac{4}{13}\right)$
(ii) Non-parametric form of vector equation of a plane
$
\vec{r} \cdot\left(\frac{12 \vec{i}+3 \vec{j}-4 \vec{k}}{13}\right)=5
$

(iii) Length of the perpendicular to the plane from the origin is 5 units.
Question 3.
Find the vector and Cartesian equation of the plane passing through the point with position vector $2 \hat{i}+6 \hat{j}+3 \hat{k}$ and normal to the vector $\hat{i}+3 \hat{j}+5 \hat{k}$ Solution:
$
\text { Given, } \vec{a}=2 \hat{i}+6 \hat{j}+3 \hat{k} \text { and } \vec{n}=\hat{i}+3 \hat{j}+5 \hat{k}
$$
Vector equation of the plane
$
\begin{aligned}
\vec{r} \cdot \vec{n} & =\vec{a} \cdot \vec{n} \\
\vec{r} \cdot \vec{n} & =(2 \hat{i}+6 \hat{j}+3 \hat{k})(\hat{i}+3 \hat{j}+5 \hat{k}) \\
\vec{r} \cdot(\vec{i}+3 \vec{j}+5 \vec{k}) & =2+18+15 \Rightarrow \vec{r} \cdot(\vec{i}+3 \vec{j}+5 \vec{k})=35
\end{aligned}
$
Cartesian equation of the plane
$
\begin{aligned}
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(\vec{i}+3 \vec{j}+5 \vec{k}) & =35 \\
x+3 y+5 z & =35
\end{aligned}
$
Question 4.
A plane passes through the point $(-1,1,2)$ and the normal to the plane of magnitude $3 \sqrt{3}$ makes equal acute angles with the coordinate axes. Find the equation of the plane.
Solution:
Given magnitude $=3 \sqrt{3}$ and $\vec{a}=-\vec{i}+\vec{j}+2 \vec{k}$
Then, the normal vector makes equal acute angle with the coordinate axes.

We know that $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$ (But $\alpha=\beta=\gamma$ )
$
\begin{aligned}
3 \cos ^2 \alpha & =1 \quad \Rightarrow \cos ^2 \alpha=1 / 3 \\
\cos \alpha & =\frac{1}{\sqrt{3}} \\
\therefore \quad \text { so } \vec{n} & =3 \sqrt{3}\left[\frac{1}{\sqrt{3}} \vec{i}+\frac{1}{\sqrt{3}} \vec{j}+\frac{1}{\sqrt{3}} \vec{k}\right] \\
\vec{n} & =3 \vec{i}+3 \vec{j}+3 \vec{k}
\end{aligned}
$
Vector equation of the plane
$
\begin{aligned}
\vec{r} \cdot \vec{n} & =\vec{a} \cdot \vec{n} \\
\vec{r} \cdot \vec{n} & =(-\vec{i}+\vec{j}+2 \vec{k})(3 \vec{i}+3 \vec{j}+3 \vec{k}) \\
\vec{r} \cdot(3 \hat{i}+3 \hat{j}+3 \hat{k}) & =-3+3+6 \\
\vec{r} \cdot(3 \vec{i}+3 \vec{j}+3 \vec{k}) & =6 \text { (or }) \vec{r} \cdot(\vec{i}+\vec{j}+\vec{k})=2
\end{aligned}
$
Cartesian equation of the plane
$
\begin{aligned}
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+3 \vec{j}+3 \vec{k}) & =6 \\
\text { (or) } \quad x+y+z & =2
\end{aligned} \quad \Rightarrow 3 x+3 y+3 z=6
$
Question 5.
Find the intercepts cut off by the plane $\vec{r} \cdot(6 \hat{i}+4 \hat{j}-3 \hat{k})=12$ on the coordinate axes.

Solution:
$
\vec{r} \cdot(6 \vec{i}+4 \vec{j}-3 \vec{k})=12
$
Compare the above equations into $\vec{r} \cdot \vec{n}=q$ so $q=12$
Let $a, b, c$ are intercepts of $x$-axis, $y$-axis and $z$-axis respectively.
Clearly

$\begin{array}{l|c|c}
\frac{q}{a}=16 & \frac{q}{b}=4 & \frac{q}{c}=-3 \\
\frac{12}{6}=a & \frac{12}{4}=b & \frac{12}{-3}=c \\
2=a & 3=b & -4=c
\end{array}$

$\mathrm{x}$-intercept $=2 ; y$-intercept $=3 ; z$-intercept $=-4$
Question 6.
If a plane meets the coordinate axes at $\mathrm{A}, \mathrm{B}, \mathrm{C}$ such that the centroid of the triangle $\mathrm{ABC}$ is the point $(\mathrm{u}, \mathrm{v}$, w), find the equation of the plane.
Solution:
Let $\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0, \mathrm{~b}, 0), \mathrm{C}(0,0, \mathrm{c})$
centroid of $\triangle \mathrm{ABC}=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$
Given centroid $=(u, v, w)$
Equate these two centroids we get
$
\begin{array}{l|l|l}
\frac{a}{3}=u & \frac{b}{3}=v & \frac{c}{3}=w \\
a=3 u & b=3 v & c=3 w
\end{array}
$
We know that intercept form the plane is

$
\begin{aligned}
& \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \\
& \frac{x}{u}+\frac{y}{v}+\frac{z}{w}=3
\end{aligned} \quad \Rightarrow \frac{x}{3 u}+\frac{y}{3 v}+\frac{z}{3 w}=1
$