Exercise 6.8 - 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 $6.8$
Question 1.
Show that the straight lines $\vec{r}=(5 \hat{i}+7 \hat{j}-3 \hat{k})+s(4 \hat{i}+4 \hat{j}-5 \hat{k})$ and $\vec{r}=(8 \hat{i}+4 \hat{j}+5 \hat{k})+t(7 \hat{i}+\hat{j}+3 \hat{k})$ are coplanar. Find the vector equation of the plane in which they lie.
Solution:
Let $\vec{a}=5 \vec{i}+7 \vec{j}-3 \vec{k}$ and $\vec{b}=4 \vec{i}+4 \vec{j}-5 \vec{k}$
$
\vec{c}=8 \vec{i}+4 \vec{j}+5 \vec{k} \text { and } \vec{d}=7 \vec{i}+\vec{j}+3 \vec{k}
$
We know that given two lines are coplanar if
$
\begin{aligned}
(\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d}) & =0 \\
\vec{b} \times \vec{d} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
4 & 4 & -5 \\
7 & 1 & 3
\end{array}\right|=\vec{i}(12+5)-\vec{j}(12+35)+\vec{k}(4-28) \\
\vec{b} \times \vec{d} & =17 \vec{i}-47 \vec{j}-24 \vec{k} \\
\vec{c}-\vec{a} & =(8 \vec{i}+4 \vec{j}+5 \vec{k})-(5 \vec{i}+7 \vec{j}-3 \vec{k})=3 \vec{i}-3 \vec{j}+8 \vec{k}
\end{aligned}
$
$
\text { (1) } \Rightarrow(3 \vec{i}-3 \vec{j}+8 \vec{k}) \cdot(17 \vec{i}-47 \vec{j}-24 \vec{k})=51+141-192=0
$
$\therefore$ The two given lines are coplanar so, the non-parametric vector equation is
$
\begin{aligned}
(\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{d}) & =0 \\
\vec{r} \cdot(\vec{b} \times \vec{d}) & =\vec{a} \cdot(\vec{b} \times \vec{d}) \\
\vec{r} \cdot(17 \vec{i}-47 \vec{j}-24 \vec{k}) & =(5 \vec{i}+7 \vec{j}-3 \vec{k})(17 \vec{i}-47 \vec{j}-24 \vec{k}) \\
\vec{r} \cdot(17 \vec{i}-47 \vec{j}-24 \vec{k}) & =85-329+72 \quad \Rightarrow \vec{r} \cdot(17 \vec{i}-47 \vec{j}-24 \vec{k})=-172
\end{aligned}
$

Question 2.
Show that lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{3}$ and $\frac{x-1}{-3}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar. Also, find the plane containing these lines.
Solution:
From the lines we have, $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(2,3,4)$ and $\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(1,4,5)$
$\left(\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3\right)=(1,1,3)$ and $\left(\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3\right)=(-3,2,1)$
Condition for coplanarity
$
\begin{aligned}
\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
b_1 & b_2 & b_3 \\
d_1 & d_2 & d_3
\end{array}\right| & =0 \\
& =\left|\begin{array}{rrr}
-1 & 1 & 1 \\
1 & 1 & 3 \\
-3 & 2 & 1
\end{array}\right|=-(1-6)-1(1+9)+1(2+3) \\
& =5-10+5=0
\end{aligned}
$
$\therefore$ The given two lines are coplanar.
Cartesian form of equation of the plane containing the two given coplanar lines.
$
\begin{aligned}
& \left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
b_1 & b_2 & b_3 \\
d_1 & d_2 & d_3
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
x-2 & y-3 & z-4 \\
1 & 1 & 3 \\
-3 & 2 & 1
\end{array}\right|=0 \\
& (x-2)[1-6]-(y-3)[1+9]+(z-4)[2+3]=0 \\
& -5(x-2)-10(y-3)+5(z-4)=0 \\
& -5 x+10-10 y+30+5 z-20=0 \\
& -5 x-10 y+5 z+20=0 \\
& (\div 6 y-5) \Rightarrow x+2 y-2 z-4=0
\end{aligned}
$

Question 3.
If the straight lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{m^2}$ and $\frac{x-3}{1}=\frac{y-2}{m^2}=\frac{z-1}{2}$ are coplanar, find the district real values of $m$.
Solution:
From the lines we have, $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(1,2,3)$ and $\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(3,2,1)$
$\left(b_1, b_2, b_3\right)=\left(1,2, m^2\right)$ and $\left(d_1, d_2, d_3\right)=\left(1, m^2, 2\right)$
Condition for coplanarity
$
\begin{aligned}
& \left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
b_1 & b_2 & b_3 \\
d_1 & d_2 & d_3
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
2 & 0 & -2 \\
1 & 2 & m^2 \\
1 & m^2 & 2
\end{array}\right|=0 \\
& 2\left[4-m^4\right]-2\left[m^2-2\right]=0 \\
& (\div \text { by } 2) \Rightarrow 4-m^4-m^2+2=0 \quad \Rightarrow m^4+m^2-6=0 \\
& \left(m^2+3\right)\left(m^2-2\right)=0 \\
& \begin{array}{c|c}
m^2+3=0 & m^2-2=0 \\
m^2=-3 & m^2=2 \\
\text { (not possible) } & m=\pm \sqrt{2}
\end{array} \\
&
\end{aligned}
$

Question 4.
If the straight lines $\frac{x-1}{2}=\frac{y+1}{\lambda}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{\lambda}$ : are coplanar, find $\lambda$ and equation of the planes containing these two lines.
Solution:
From the lines we have, $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(1,-1,0)$ and $\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(-1,-1,0)$
$\left(\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3\right)=(2, \lambda, 2)$ and $\left(\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3\right)=(5,2, \lambda)$
Condition for coplanarity

$\begin{aligned}
& -2\left(\lambda^2-4\right)=0 \quad \Rightarrow \lambda^2=4 \\
& \lambda=\pm 2 \\
& \text { (i) If } \lambda=2 \\
& \left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
b_1 & b_2 & b_3 \\
d_1 & d_2 & d_3
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
x-1 & y+1 & z \\
2 & 2 & 2 \\
5 & 2 & 2
\end{array}\right|=0 \\
& (x-1)[0]-(y+1)[4-10]+z[4-10]=0 \\
& 6(y+1)-6(z)=0 \\
& 6 y+6-6 z=0 \\
& (\div \text { by } 6) \Rightarrow(y-z+1)=0 \\
& \left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
b_1 & b_2 & b_3 \\
d_1 & d_2 & d_3
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
x-1 & y+1 & z \\
2 & -2 & 2 \\
5 & 2 & -2
\end{array}\right|=0 \\
& (\mathrm{x}-1)[0]-(\mathrm{y}+1)[-4-10]+\mathrm{z}[4+10]=0 \\
& 14(y+1)+14 z=0 \Rightarrow 14 y+14+14 z=0 \\
& (\div \text { by } 14) \Rightarrow y+z+1=0 \\
&
\end{aligned}$