Exercise 6.7 - 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 $6.7$
Question 1.
Find the non-parametric form of vector equation, and Cartesian equation of the plane passing through the point $(2,3,6)$ and parallel to the straight lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-3}{1}$ and $\frac{x+3}{2}=\frac{y-3}{-5}=\frac{z+1}{-3}$
Solution:
The required plane passing through the point $\vec{a}=2 \vec{i}+3 \vec{j}+6 \vec{k}$ and parallel to the vectors $\vec{b}=2 \vec{i}+3 \vec{j}+\vec{k}$ and $\vec{c}=2 \vec{i}-5 \vec{j}-3 \vec{k}$
$
\begin{aligned}
& \vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 3 & 1 \\
2 & -5 & -3
\end{array}\right|=\vec{i}(-9+5)-\vec{j}(-6-2)+\vec{k}(-10-6) \\
& \vec{b} \times \vec{c}=-4 \vec{i}+8 \vec{j}-16 \vec{k}
\end{aligned}
$
Non-parametric form of vector equation
$
\begin{array}{rlrl}
(\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{c}) & =0 & \text { (or) } \vec{r} \cdot(\vec{b} \times \vec{c})=\vec{a} \cdot(\vec{b} \times \vec{c}) \\
\vec{r}(-4 \vec{i}+8 \vec{j}-16 \vec{k}) & =(2 \vec{i}+3 \vec{j}+6 \vec{k}) \cdot(-4 \vec{i}+8 \vec{j}-16 \vec{k}) \\
\vec{r} \cdot(-4 \vec{i}+8 \vec{j}-16 \vec{k}) & =-8+24-96 & \\
\vec{r} \cdot(-4 \vec{i}+8 \vec{j}-16 \vec{k}) & =-80 & \Rightarrow-4[\vec{r} \cdot(\vec{i}-2 \vec{j}+4 \vec{k})]=-80 \\
\vec{r} \cdot(\vec{i}-2 \vec{j}+4 \vec{k}) & =20 & &
\end{array}
$
Cartesian equation
$
\begin{aligned}
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(\vec{i}-2 \vec{j}+4 \vec{k}) & =20 \\
x-2 y+4 z & =20
\end{aligned}
$
Question 2.
Find the parametric form of vector equation, and Cartesian equations of the plane passing through the points $(2,2,1),(9,3,6)$ and perpendicular to the plane $2 x+6 y+6 z=9$.
Solution:

The required plane passes through the points
$
\vec{a}=2 \vec{i}+2 \vec{j}+\vec{k}, \vec{b}=9 \vec{i}+3 \vec{j}+6 \vec{k}
$
and parallel to the vector $\vec{c}=2 \vec{i}+6 \vec{j}+6 \vec{k}$
$
\vec{b}-\vec{a}=(9 \vec{i}+3 \vec{j}+6 \vec{k})-(2 \vec{i}+2 \vec{j}+\vec{k})=7 \vec{i}+\vec{j}+5 \vec{k}
$
Non-parametric form of vector equation
$
\begin{aligned}
(\vec{r}-\vec{a}) \cdot((\vec{b}-\vec{a}) \times \vec{c}) & =0 \\
(\vec{b}-\vec{a}) \times \vec{c} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
7 & 1 & 5 \\
2 & 6 & 6
\end{array}\right|=\vec{i}(6-30)-\vec{j}(42-10)+\vec{k}(42-2) \\
& =-24 \hat{i}-32 \hat{j}+40 \hat{k}
\end{aligned}
$
$
\text { (1) } \Rightarrow \quad \begin{aligned}
(\vec{r}-\vec{a}) \cdot(-24 \hat{i}-32 \hat{j}+40 \hat{k}) & =0 \\
\vec{r} \cdot(-24 \hat{i}-32 \hat{j}+40 \hat{k}) & =\vec{a} \cdot(-24 \vec{i}-32 \vec{j}+40 \vec{k}) \\
\vec{r} \cdot(24 \vec{i}-32 \vec{j}+40 \vec{k}) & =(2 \vec{i}+2 \vec{j}+\vec{k})(-24 \vec{i}-32 \vec{j}+40 \vec{k}) \\
\vec{r} \cdot(-24 \hat{i}-32 \hat{j}+40 \hat{k}) & =-48-64+40 \\
-8[\vec{r} \cdot(3 \hat{i}+4 \hat{j}-5 \hat{k})] & =-72 \\
\vec{r} \cdot(3 \hat{i}+4 \hat{j}-5 \hat{k}) & =9
\end{aligned}
$
Cartesian equation
$
\begin{array}{r}
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+4 \vec{j}-5 \vec{k})=9 \\
3 x+4 y-5 z-9=0
\end{array}
$
Question 3.
Find the parametric form of vector equation and Cartesian equations of the plane passing through the points $(2,2,1),(1,-2,3)$ and parallel to the straight line passing through the points $(2,1,-3)$ and $(-1,5,-8)$.
Solution:
Equation of the straight line passing through the points $(2,1,-3)$ and $(-1,5,-8)$ is

$
\frac{x-2}{-3}=\frac{y-1}{4}=\frac{z+3}{-5}
$
$\therefore$ The required plane passing through the points
$
\vec{a}=2 \vec{i}+2 \vec{j}+\vec{k} \text { and } \vec{b}=\vec{i}-2 \vec{j}+3 \vec{k}
$
and parallel to $\vec{c}=-3 \vec{i}+4 \vec{j}-5 \vec{k}$
Parametric form of vector equation
$
\begin{aligned}
& \vec{r}=\vec{a}+s(\vec{b}-\vec{a})+t \vec{c} \\
& \vec{r}=(2 \vec{i}+2 \vec{j}+\vec{k})+s(\vec{i}-4 \vec{j}+2 \vec{k})+t(-3 \vec{i}+4 \vec{j}-5 \vec{k}), s, t \in \mathbb{R}
\end{aligned}
$
Cartesian equation:
$
\begin{aligned}
& \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 \\
c_1 & c_2 & c_3
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
x-2 & y-2 & z-1 \\
-1 & -4 & 2 \\
-3 & 4 & -5
\end{array}\right|=0 \\
& (\mathrm{x}-2)[20-8]-(\mathrm{y}-2)[5+6]+(\mathrm{z}-1)[-4-12]=0 \\
& 12(\mathrm{x}-2)-11(\mathrm{y}-2)-16(\mathrm{z}-1)=0 \\
& 12 \mathrm{x}-24-11 \mathrm{y}+22-16 \mathrm{z}+16=0 \\
& 12 \mathrm{x}-11 \mathrm{y}-16 \mathrm{z}+14=0
\end{aligned}
$

Question 4.
Find the non-parametric form of vector equation of the plane passing through the point $(1,-2,4)$ and perpendicular to the plane $\mathrm{x}+2 \mathrm{y}-3 \mathrm{z}=11$ and parallel to the line $\frac{x+7}{3}=\frac{y+3}{-1}=\frac{z}{1}$
Solution:
The required plane passing through the point $\vec{a}=\vec{i}-2 \vec{j}+4 \vec{k}$ and parallel to the plane $\vec{b}=\vec{i}+2 \vec{j}-3 \vec{k}$

and parallel to the line $\vec{c}=3 \vec{i}-\vec{j}+\vec{k}$
$
\begin{aligned}
& \vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & 2 & -3 \\
3 & -1 & 1
\end{array}\right|=\vec{i}(2-3)-\vec{j}(1+9)+\vec{k}(-1-6) \\
& \vec{b} \times \vec{c}=-\vec{i}-10 \vec{j}-7 \vec{k}
\end{aligned}
$
Non-parametric form of vector equation
$
\begin{aligned}
& (\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{c})=0 \\
& (\vec{r}-\vec{a}) \cdot(-\vec{i}-10 \vec{j}-7 \vec{k})=0 \\
& \vec{r} \cdot(-\vec{i}-10 \vec{j}-7 \vec{k})=\vec{a} \cdot(-\vec{i}-10 \vec{j}-7 \vec{k}) \\
& \vec{r} \cdot(-\vec{i}-10 \vec{j}-7 \vec{k})=(\vec{i}-2 \vec{j}+4 \vec{k})(-\vec{i}-10 \vec{j}-7 \vec{k}) \\
& \vec{r} \cdot(-\vec{i}-10 \vec{j}-7 \vec{k})=-1+20-28 \\
& \vec{r} \cdot(-\vec{i}-10 \vec{j}-7 \vec{k})=-9 \quad \text { (or) } \vec{r} \cdot(\vec{i}+10 \vec{j}+7 \vec{k})=9 \\
&
\end{aligned}
$
Cartesian equation
$
\begin{array}{r}
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(\vec{i}+10 \vec{j}+7 \vec{k})=9 \\
x+10 y+7 z-9=0
\end{array}
$

Question 5.
Find the parametric form of vector equation, and Cartesian equations of the plane containing the line $\vec{r}=(\hat{i}-\hat{j}+3 \hat{k})+t(2 \hat{i}-\hat{j}+4 \hat{k})$ and perpendicular to the plane $\vec{r} \cdot(\hat{i}+2 \hat{j}+\hat{k})=8$.
Solution:
The required plane passing through the point $\vec{a}=\vec{i}-\vec{j}+3 \vec{k}$ and parallel to $\vec{b}=2 \vec{i}-\vec{j}+4 \vec{k}$ and $\vec{c}=\vec{i}+2 \vec{j}+\vec{k}$
Parametric form of vector equation
$
\begin{aligned}
& \vec{r}=\vec{a}+s \vec{b}+t \vec{c} \\
& \vec{r}=(\vec{i}-\vec{j}+3 \vec{k})+s(2 \vec{i}-\vec{j}+4 \vec{k})+t(\vec{i}+2 \vec{j}+\vec{k}), s, t \in \mathrm{R}
\end{aligned}
$
Cartesian equation
$
\begin{aligned}
\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right| & =0 \\
\left|\begin{array}{ccc}
x-1 & y+1 & z-3 \\
2 & -1 & 4 \\
1 & 2 & 1
\end{array}\right| & =0
\end{aligned}
$
$
\begin{aligned}
& (x-1)[-1-8]-(y+1)[z-4]+(z-3)[4+1]=0 \\
& -9(x-1)+2(y+1)+5(z-3)=0 \\
& -9 x+9+2 y+2+5 z-15=0 \\
& -9 x+2 y+5 z-4=0 \\
& 9 x-2 y-5 z+4=0
\end{aligned}
$

Question 6.
Find the parametric vector, non-parametric vector and Cartesian form of the equation of the plane passing through the point $(3,6,-2),(-1,-2,6)$ and $(6,4,-2)$.
Solution:
The required plane passing through the points
$
\begin{aligned}
\vec{a} & =3 \vec{i}+6 \vec{j}-2 \vec{k} & & \\
\vec{b} & =-\vec{i}-2 \vec{j}+6 \vec{k} & & \text { and } \vec{c}=6 \vec{i}+4 \vec{j}-2 \vec{k} \\
\vec{b}-\vec{a} & =-4 \vec{i}-8 \vec{j}+8 \vec{k} & & \text { and } \vec{c}-\vec{a}=3 \vec{i}-2 \vec{j}
\end{aligned}
$
Parametric form of vector equation
$
\begin{aligned}
& \vec{r}=\vec{a}+s(\vec{b}-\vec{a})+t(\vec{c}-\vec{a}) \\
& \vec{r}=(3 \vec{i}+6 \vec{j}-2 \vec{k})+s(-4 \vec{i}-8 \vec{j}+8 \vec{k})+t(3 \vec{i}-2 \vec{j}), s, t \in \mathbb{R}
\end{aligned}
$
Non-parametric form of vector equation
$
\begin{aligned}
& (\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-4 & -8 & 8 \\
3 & -2 & 0
\end{array}\right|=\vec{i}(0+16)-\vec{j}(0-24)+\vec{k}(8+24) \\
& =16 \vec{i}+24 \vec{j}+32 \vec{k} \\
& \therefore \quad(\vec{r}-\vec{a}) \cdot((\vec{b}-\vec{a}) \times(\vec{c}-\vec{a}))=0 \\
& (\vec{r}-\vec{a}) \cdot(16 \vec{i}+24 \vec{j}+32 \vec{k})=0 \\
& \vec{r} \cdot(16 \vec{i}+24 \vec{j}+32 \vec{k})=\vec{a} \cdot(16 \vec{i}+24 \vec{j}+32 \vec{k}) \\
& \vec{r} \cdot(16 \vec{i}+24 \vec{j}+32 \vec{k})=(3 \vec{i}+6 \vec{j}-2 \vec{k})(16 \vec{i}+24 \vec{j}+32 \vec{k}) \\
& \vec{r} \cdot(16 \vec{i}+24 \vec{j}+32 \vec{k})=48+144-64 \\
& \vec{r} \cdot(16 \vec{i}+24 \vec{j}+32 \vec{k})=128 \quad \Rightarrow 8[\vec{r} \cdot(2 \vec{i}+3 \vec{j}+4 \vec{k})]=128 \\
& \vec{r} \cdot(2 \vec{i}+3 \vec{j}+4 \vec{k})=16 \\
&
\end{aligned}
$
Cartesian equation
$
\begin{aligned}
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(2 \vec{i}+3 \vec{j}+4 \vec{k}) & =16 \\
2 x+3 y+4 z-16 & =0
\end{aligned}
$

Question 7.
Find the non-parametric form of vector equation, and Cartesian equations of the plane
$
\vec{r}=(6 \hat{i}-\hat{j}+\hat{k})+s(-\hat{i}+2 \hat{j}+\hat{k})+t(-5 \hat{i}-4 \hat{j}-5 \hat{k})
$
Solution:
Given $\vec{a}=6 \hat{i}-\hat{j}+\hat{k}, \vec{b}=-\vec{i}+2 \vec{j}+\vec{k}$ and $\vec{c}=-5 \vec{i}-4 j-5 \vec{k}$
$
\begin{aligned}
& \vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-1 & 2 & 1 \\
-5 & -4 & -5
\end{array}\right|=\vec{i}(-10+4)-\vec{j}(5+5)+\vec{k}(4+10) \\
& \vec{b} \times \vec{c}=-6 \vec{i}-10 \vec{j}+14 \vec{k}
\end{aligned}
$
Non-parametric form of vector equation
$
\begin{aligned}
&(\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{c})=0 \\
&(\vec{r}-\vec{a}) \cdot(-6 \vec{i}-10 \vec{j}+14 \vec{k})=0 \\
& {[\div \mathrm{by}-2] } \\
&(\vec{r}-\vec{a}) \cdot(3 \vec{i}+5 \vec{j}-7 \vec{k})=0 \\
& \vec{r} \cdot(3 i+5 \vec{j}-7 \vec{k})=\vec{a} \cdot(3 \vec{i}+5 \vec{j}-7 \vec{k}) \\
& \vec{r} \cdot(3 i+5 \vec{j}-7 \vec{k})=(6 \vec{i}-\vec{j}+\vec{k})(3 \vec{i}+5 \vec{j}-7 \vec{k}) \\
& \vec{r} \cdot(3 \vec{i}+5 \vec{j}-7 \vec{k})=18-5-7 \quad \Rightarrow \vec{r} \cdot(3 \vec{i}+5 \vec{j}-7 \vec{k})=6
\end{aligned}
$
Cartesian equation
$
\begin{aligned}
(x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+5 \vec{j}-7 \vec{k}) & =6 \\
3 x+5 y-7 z-6 & =0
\end{aligned}
$