Exercise 6.6-Additional Problems - 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

Additional Problems
Question 1.
Find the vector and cartesian equations of a plane which is at a distance of 18 units from the origin and which is normal to the vector $2 \vec{i}+7 \vec{j}+8 \vec{k}$
Solution:
$
\begin{aligned}
\hat{n} & =\frac{\vec{n}}{|\vec{n}|}=\frac{2 \vec{i}+7 \vec{j}+8 \vec{k}}{\sqrt{4+49+64}}=\frac{2 \vec{i}+7 \vec{j}+8 \vec{k}}{\sqrt{117}} \\
& =\frac{2 \vec{i}+7 \vec{j}+8 \vec{k}}{3 \sqrt{13}}
\end{aligned}
$
Hence, the required vector equation of the plane is $\vec{r} \cdot \hat{n}=p$.
i.e.
$
\frac{\vec{r} \cdot(2 \vec{i}+7 \vec{j}+8 \vec{k})}{3 \sqrt{13}}=18 \text { i.e. } \vec{r} \cdot(2 \vec{i}+7 \vec{j}+8 \vec{k})=54 \sqrt{13}
$
Taking $\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}$ we get the cartesian equation as
$
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(2 \vec{i}+7 \vec{j}+8 \vec{k})=54 \sqrt{13}
$
i.e.
$
2 x+7 y+8 z=54 \sqrt{13}
$

Question 2.
Find the unit vector to the plane $2 x-y+2 z=5$.
Solution:
Writing the plane in normal form we get,
$
\begin{aligned}
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(2 \vec{i}-\vec{j}+2 \vec{k}) & =5 \text { i.e. } \vec{r} \cdot \vec{n}=\vec{p} \\
\text { Here normal vector } & =\vec{n}=2 \vec{i}-\vec{j}+2 \vec{k} \\
|\vec{n}| & =\sqrt{4+1+4}=3
\end{aligned}
$
So, the unit vector normal to the plane $=\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\pm \frac{2 \vec{i}-\vec{j}+2 \vec{k}}{3}$

Question 3.
Find the length of the perpendicular from the origin to the plane $\vec{r} \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26$.
Solution:
Taking the equation of the plane in cartesian form we get,
$(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26$
i.e., $3 x+4 y+12 z-26=0$
The length of the perpendicular from $(0,0,0)$ to the above plane is $\pm \frac{-26}{\sqrt{9+16+144}}=\frac{+26}{13}=2$ units
Question 4.
The foot of the perpendicular drawn from the origin to a plane is $(8,-4,3)$. Find the equation of the plane.

Solution:
The required plane passes through the point $\mathrm{A}(8,-4,3)$ and is perpendicular to $\overrightarrow{\mathrm{OA}}$.
$\therefore \quad \vec{a}=8 \vec{i}-4 \vec{j}+3 \vec{k}$ and $\vec{n}=\overrightarrow{\mathrm{OA}}=8 \vec{i}-4 \vec{j}+3 \vec{k}$
The required equation of the plane is $\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}$
$
\vec{r} \cdot(8 \vec{i}-4 \vec{j}+3 \vec{k})=(8 \vec{i}-4 \vec{j}+3 \vec{k}) \cdot(8 \vec{i}-4 \vec{j}+3 \vec{k})=64+16+9=89
$
The cartesian equation is $8 x-4 y+3 z=89$.
Question 5.
Find the equation of the plane through the point whose position vector is $2 \vec{i}-\vec{j}+\vec{k}$ and perpendicular to the vector $4 \vec{i}+2 \vec{j}-3 \vec{k}$.
Solution:
The required plane is perpendicular to $4 \vec{i}+2 \vec{j}-3 \vec{k}$
So, it is parallel to the plane $4 x+2 y-3 z=k$
$\therefore$ the equation of the plane is $4 \mathrm{x}+2 \mathrm{y}-3 \mathrm{z}=\mathrm{k}$
The plane passes through the point $(2,-1,1)$
$\Rightarrow(4)(2)+2(-1)-3(1)=\lambda$ i.e. $\lambda=8-2-3=3$
So, the equation of the plane is $4 x+2 y-3 z=3$.

Question 6.
Find the vector and cartesian equations of the plane passing through the point $(2,-1,4)$ and parallel to the plane $\vec{r} \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})=7$
Solution:
The given plane is $\vec{r} \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})=7$
i e. $(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})=7$
i.e. $4 x-12 y-3 z=1$
The required plane is parallel to the above plane. So, the equation of the required plane is $4 x-12 y-3 z-k$. The plane passes through $(2,-1,4)$.
$\Rightarrow 4(2)-12(-1)-3(4)=\mathrm{k}$ i.e. $\mathrm{k}=8+12-12=8$
So, the equation of the plane is $4 x-12 y-3 z=8$.