Exercise 6.7-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 Problem
Question 1.
Find the vector and cartesian equations of the plane containing the line $\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{3}$ and parallel to the line $\frac{x+1}{3}=\frac{y-1}{2}=\frac{z+1}{1}$
Solution:
The required plane passes through the point $\mathrm{A}(2,2,1)$ and parallel to
$
\vec{u}=2 \vec{i}+3 j+3 \vec{k} \text { and } \vec{v}=3 \vec{i}+2 \vec{j}+\vec{k}
$
The required equation is $\vec{r}=\vec{a}+s \vec{u}+t \vec{v}$
i.e., $\quad x \vec{i}+y \vec{j}+z \vec{k}=(2 \vec{i}+2 \vec{j}+\vec{k})+s(2 \vec{i}+3 \vec{j}+3 \vec{k})+t(3 \vec{i}+2 \vec{j}+\vec{k})$
Cartesian form: The equation of the plane is
$
\begin{aligned}
& \left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|=0 \quad \text { i.e. }\left|\begin{array}{ccc}
x-2 & y-2 & z-1 \\
2 & 3 & 3 \\
3 & 2 & 1
\end{array}\right|=0 \\
& \text { i.e., }(x-2)[3-6]-(y-2)[2-9]+(z-1)[4-9]=0 \\
& \text { i.e., }(x-2)(-3)+(y-2)(7)-5(z-1)=0 \\
& -3 x+6+7 y-14-5 z+5=0 \\
& -3 x+7 y-5 z-3=0 \text { i.e. } 3 x-7 y+5 z+3=0 \\
&
\end{aligned}
$

Question 2.
Find the vector and cartesian equation of the plane passing through the point $(1,3,2)$ and parallel to the lines $\frac{x+1}{2}=\frac{y+2}{-1}=\frac{z+3}{3}$ and $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z+2}{2}$.
Solution:
The required plane passes through the point $\mathrm{A}=(1,3,2)$ and parallel to the vectors
$
\vec{u}=2 \vec{i}-\vec{j}+3 \vec{k} \text { and } \vec{v}=\vec{i}+2 \vec{j}+2 \vec{k}
$
The required equation is $\vec{r}=\vec{a}+s \vec{\mu}+t \vec{v}$
i.e. $\quad x \vec{i}+y \vec{j}+z \vec{k}=(\vec{i}+3 \vec{j}+2 \vec{k})+s(2 \vec{i}-\vec{j}+3 \vec{k})+t(\vec{i}+2 \vec{j}+2 \vec{k})$
Cartesian equation of the plane is $\left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=0$
So, the equation of the plane is $\left|\begin{array}{ccc}x-1 & y-3 & z-2 \\ 2 & -1 & 3 \\ 1 & 2 & 2\end{array}\right|=0$
i.e. $(x-1)(-2-6)-(y-3)(4-3)+(z-2)(4+1)=0$
$-8(\mathrm{x}-1)-1(\mathrm{y}-3)+5(\mathrm{z}-2)=0$
$-8 \mathrm{x}+8-\mathrm{y}+3+5 \mathrm{z}-10=0$
$-8 x-y+5 z+1=0$ i.e. $8 x+y-5 z-1=0$

Question 3.
Find the vector and cartesian equations of the plane passing through the point $(-1,3,2)$ and perpendicular to the planes $x+2 y+2 z=5$ and $3 x+y+2 z=8$
Solution:
The normal vector to the given planes $\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}=5$ and $3 \mathrm{x}+\mathrm{y}+2 \mathrm{z}=8$ are respectively $\vec{i}+2 \vec{j}+2 \vec{k}$ and $3 \vec{i}+\vec{j}+2 \vec{k}$. These vectors are parallel to the required plane
The required plane passes through the point $\mathrm{A}(-1,3,2)$ and parallel to the vectors
$
\vec{u}=\vec{i}+2 \vec{j}+2 \vec{k} \text { and } \vec{v}=3 \vec{i}+\vec{j}+2 \vec{k}
$
The required equation is $\vec{r}=\vec{a}+s \vec{u}+t \vec{v}$
i.e.
$
\vec{r}=-\vec{i}+3 \vec{j}+2 \vec{k}+s(\vec{i}+2 \vec{j}+2 \vec{k})+t(3 \vec{i}+\vec{j}+2 \vec{k})
$
Cartesian form is $\left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=0$
So, the equation of the plane is $\left|\begin{array}{ccc}x+1 & y-3 & z-2 \\ 1 & 2 & 2 \\ 3 & 1 & 2\end{array}\right|=0$
i.e., $(x+1)(4-2)-(y-3)(2-6)+(z-2)(1-6)=0$
$2(\mathrm{x}+1)+4(\mathrm{y}-3)-5(\mathrm{z}-2)=0$
$2 \mathrm{x}+2+4 \mathrm{y}-12-5 \mathrm{z}+10=0$
$2 x+4 y-5 z=0$

Question 4.
Find the vector and cartesian equations of the plane passing through the points $\mathrm{A}(1,-2,3)$ and $\mathrm{B}(-1,2,-1)$ and is parallel to the line $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z-1}{4}$
Solution:
The vector equation of the plane is
$\vec{r}=(1-s) \vec{a}+s \vec{b}+t \vec{c}$ where $s$ and $t$ are scalars
$\therefore \quad \vec{r}=(1-s)(\vec{i}-2 \vec{j}+3 \vec{k})+s(-\vec{i}+2 \vec{j}-\vec{k})+t(2 \vec{i}+3 \vec{j}+4 \vec{k})$
The cartesian equation of the plane is $\left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l & m & n\end{array}\right|=0$
So, the equation of the plane is $\left|\begin{array}{ccc}x-1 & y+2 & z-3 \\ -2 & 4 & -4 \\ 2 & 3 & 4\end{array}\right|=0$
i.e, $(x-1)[16+12]-(y+2)(-8+8)+(z-3)(-6-8)=0$
$28(x-1)-14(z-3)=0$
$28 \mathrm{x}-28-14 \mathrm{z}+42=0$
$28 \mathrm{x}-14 \mathrm{z}+14=0$
$(\div$ by 14$) \Rightarrow 2 x-z+1=0$

Question 5.
Find the vector and cartesian equations of the plane through the points $(1,2,3)$ and $(2,3,1)$ and perpendicular to the plane $3 x-2 y+4 z-5=0$.
Solution:
The vector normal to the plane $3 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}-5=0$ is $3 \vec{i}-2 \vec{j}+4 \vec{k}$
The required plane is parallel to the vector $3 \vec{i}-2 \vec{j}+4 \vec{k}$
$\therefore$ Vector equation of the plane passing through two points and parallel to one vector is
$
\vec{r}=(1-s) \vec{a}+s \vec{b}+t \vec{c} \text { where } s \text { and } t \text { are scalars. }
$
So, the vector equation of the required plane is
$
\vec{r}=(1-s)(\vec{i}+2 \vec{j}+3 \vec{k})+s(2 \vec{i}+3 \vec{j}+\vec{k})+t(3 \vec{i}-2 \vec{j}+4 \vec{k})
$
The cartesian equation is $\left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l & m & n\end{array}\right|=0$
So, the equation of the plane is $\left|\begin{array}{ccc}x-1 & y-2 & z-3 \\ 1 & 1 & -2 \\ 3 & -2 & 4\end{array}\right|=0$
i.e. $(x-1)[4-4]-(y-2)[4+6]+(z-3)[-2-3]=0$
$-10(y-2)-5(z-3)=0$
$-10 y+20-5 z+15=0$
$-10 y-5 z+35=0$
$10 y+5 z-35=0$
$(\div$ by 5$) \Rightarrow 2 y+z-1=0$

Question 6.
Find the vector and cartesian equations of the plane containing the line $\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{-2}$ and passing through the point $(-1,1,-1)$.
Solution:
The required plane passes through the points $\mathrm{A}(-1,1,-1)$ and $\mathrm{B}(2,2,1)$ and parallel to the vectors $\vec{c}=2 \vec{i}+3 \vec{j}-2 \vec{k}$
The required equation is
$
\vec{r}=(1-s) \vec{a}+s \vec{b}+t \vec{c} \text { where } s \text { and } t \text { are scalars. }
$
$
\vec{r}=(1-s)(-\vec{i}+\vec{j}-\vec{k})+s(2 \vec{i}+2 \vec{j}+\vec{k})+t(2 \vec{i}+3 \vec{j}-2 \vec{k})
$
Cartesian equation of the plane is $\left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l & m & n\end{array}\right|=0$
So, the equation of the plane is $\left|\begin{array}{ccc}x+1 & y-1 & z+1 \\ 3 & 1 & 2 \\ 2 & 3 & -2\end{array}\right|=0$
i.e. $(\mathrm{x}+1)[-2-6]-(\mathrm{y}-1)[-6-4]+(\mathrm{z}+1)[9-2]=0$
$-8(\mathrm{x}+1)+10(\mathrm{y}-1)+7(\mathrm{z}+1)=0$
$-8 \mathrm{x}-8+10 \mathrm{y}-10+7 \mathrm{z}+7=0$
$-8 \mathrm{x}+10 \mathrm{y}+7 \mathrm{z}-11=0$
i.e., $8 \mathrm{x}-10 \mathrm{y}-7 \mathrm{z}+11=0$

Question 7.
Find the vector and cartesian equations of the plane passing through the points with position vectors
$3 \vec{i}+4 \vec{j}+2 \vec{k}, 2 \vec{i}-2 \vec{j}-\vec{k}$ and $7 \vec{i}+\vec{k}$
Solution:
Vector equation of the plane passing through three given non-collinear points is
$\begin{aligned}
& \vec{r}=(1-s-t) \vec{a}+s \vec{b}+t \vec{c} \text { where } s \text { and } t \text { are scalars } \\
& \therefore \quad \vec{r}=(1-s-t)(3 \vec{i}+4 \vec{j}+2 \vec{k})+s(2 \vec{i}-2 \vec{j}-\vec{k})+t(7 \vec{i}+\vec{k}) \\
& \text { Cartesian equation of the plane is }\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
x-3 & y-4 & z-2 \\
-1 & -6 & -3 \\
4 & -4 & -1
\end{array}\right|=0 \\
& \text { i.e. }(x-3)[6-12]-(y-4)[1+12]+(z-2)[4+24]=0 \\
& -6(\mathrm{x}-3)-13(\mathrm{y}-4)+28(\mathrm{z}-2)=0 \\
& -6 x+18-13 y+52+28 z-56=0 \\
& -6 x-13 y+28 z+14=0 \\
& \text { i.e. } 6 x+13 y-28 z-14=0 \\
&
\end{aligned}$

Question 8.
Derive the equation of the plane in the intercept form.
Solution:
Let the required plane makes intercepts on $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$-axes respectively as $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$.
i.e. $A=(a, 0,0)$
$\mathrm{B}=(0, \mathrm{~b}, 0)$
$\mathrm{C}=(0,0, \mathrm{c})$
The equation of the plane passing through three non-collinear points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ is

$\begin{aligned}
& \vec{r}=(1-s-t) a+s \vec{b}+t \vec{c} \text {. } \\
& \text { Here } \quad \vec{a}=a \vec{i}, \vec{b}=b \vec{j} \text { and } \vec{c}=c \vec{k} \\
& \therefore \quad \vec{r}=(1-s-t) a \vec{i}+s b \vec{j}+t c \vec{k} \\
& \text { Cartesian equation of the plane is }\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
x-a & y-0 & z-0 \\
-a & b & 0 \\
-a & 0 & c
\end{array}\right|=0 \\
& (x-a)(b c-0)-y(-a c-0)+z(a b)=0 \\
& \text { i.e. } x b c-a b c+y a c+z a b=0 \\
& x b c+y a c+z a b=a b c \\
& (\div \text { by } a b c) \Rightarrow \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \\
&
\end{aligned}$

Question 9.
Find the cartesian form of the following plane:
$
\vec{r}=(s-2 t) \vec{i}+(3-t) \vec{j}+(2 s+t) \vec{k}
$
Solution:
Now
$
\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}
$
$
(x-0) \vec{i}+(y-3) \vec{j}+(z-0) \vec{k}=(s-2 t) \vec{i}+(-t) \vec{j}+(2 s+t) \vec{k}
$
Equating $\vec{i}, \vec{j}, \vec{k}$ components and eliminating $s$ and $t$ we get,
$
\begin{aligned}
& \left|\begin{array}{ccc}
x-0 & y-3 & z-0 \\
s & 0 & 2 s \\
-2 t & -t & t
\end{array}\right|=0 \text {; i.e., }\left|\begin{array}{ccc}
x & y-3 & z \\
1 & 0 & 2 \\
-2 & -1 & 1
\end{array}\right|=0 \\
& \mathrm{x}(0+2)-(\mathrm{y}-3)(1+4)+\mathrm{z}(-1)=0 \\
& 2 \mathrm{x}-5(\mathrm{y}-3)-\mathrm{z}=0 \\
& 2 \mathrm{x}-5 \mathrm{y}+15-\mathrm{z}=0 \\
& 2 \mathrm{x}-5 \mathrm{y}-\mathrm{z}+15=0
\end{aligned}
$
which is the cartesian equation of the plane.