Exercise 6.8-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.
Show that the straight lines.
$
\begin{aligned}
& \vec{r}=(\vec{i}+\vec{j}-\vec{k})+\lambda(3 \vec{i}-\vec{j}) \\
& \vec{r}=(4 \vec{i}-\vec{k})+\mu(2 \vec{i}+3 \vec{k})
\end{aligned}
$
are coplanar. Find the vector equation of the plane in which they lie.
Solution:
Given $\vec{a}=\vec{i}+\vec{j}-\vec{k} ;$ and $\vec{b}=3 \vec{i}-\vec{j}$
$\vec{c}=4 \vec{i}-\vec{k} ;$ and $\vec{d}=2 \vec{i}+3 \vec{k}$
Conditions for coplanar
$
\begin{aligned}
(\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d}) & =0 \\
\vec{c}-\vec{a} & =(4 \vec{i}-\vec{k})-(\vec{i}+\vec{j}-\vec{k})=3 \vec{i}-\vec{j} \\
\vec{b} \times \vec{d} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & -1 & 0 \\
2 & 0 & 3
\end{array}\right|=\vec{i}(-3)-\vec{j}(9)+\vec{k}(2)=-3 \vec{i}-9 \vec{j}+2 \vec{k} \\
(\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d}) & =(3 \vec{i}-\vec{j})(-3 \vec{i}-9 \vec{j}+2 \vec{k})=-9+9+0=0 .
\end{aligned}
$
Given two lines are coplanar.
Vector equation of the plane:
$
\begin{array}{ll}
\vec{r}=\vec{a}+t \vec{b}+s \vec{d} & \Rightarrow \vec{r}=(\vec{i}+\vec{j}-\vec{k})+t(3 \vec{i}-\vec{j})+s(2 \vec{i}+3 \vec{k}) \\
\vec{r}=\vec{c}+t \vec{b}+s \vec{d} & \Rightarrow \vec{r}=(4 \vec{i}-\vec{k})+t(3 \vec{i}-\vec{j})+s(2 \vec{i}+3 \vec{k})
\end{array}
$

Question 2.
If the straight lines $\frac{x-1}{1}=\frac{y-1}{\lambda}=\frac{z-1}{1}$ and $\frac{x}{2}=\frac{y-4}{\lambda}=\frac{z-2}{3}$ are coplanar. Find $\lambda$.
Solution:
From the lines we have, $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(1,1,1)$ and $\left(\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3\right)=(1, \lambda, 1)$
$\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(0,4,2)$ and $\left(\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3\right)=(2, \lambda, 3)$
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}
-1 & 3 & 1 \\
1 & \lambda & 1 \\
2 & \lambda & 3
\end{array}\right|=0 \\
& -1(3 \lambda-\lambda)-3(3-2)+1(\lambda-2 \lambda)=0 \Rightarrow-2 \lambda-3-\lambda=0 \\
& -3 \lambda=3 \Rightarrow \lambda=-1 \\
&
\end{aligned}
$

Question 3.
If the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, then find the value of $\mathrm{k}$. Solution:
From the lines we have, $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(2,3,4)$ and $\left(\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3\right)=(1,1,-1)$ $\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(1,4,5)$ and $\left(\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3\right)=(\mathrm{k}, 2,1)$
Condition for coplanarity
$
\begin{array}{r}
\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}
-1 & 1 & 1 \\
1 & 1 & -k \\
k & 2 & 1
\end{array}\right|=0
\end{array}
$
$
-1(1+2 \mathrm{k})-1\left(1+\mathrm{k}^2\right)+1(2-\mathrm{k})=0
$

$\begin{aligned}
& -1-2 \mathrm{k}-1-\mathrm{k}^2+2-\mathrm{k}=0 \\
& -\mathrm{k}^2-3 \mathrm{k}=0 \\
& \mathrm{k}^2+3 \mathrm{k}=0 \\
& \mathrm{k}(\mathrm{k}+3)=0 \\
& \mathrm{k}=0 \text { or } \mathrm{k}=-3
\end{aligned}$

Question 4.
Show that the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar, Also find the equation of the plane containing these two lines.
Solution:
From the lines we have, $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(-3,1,5)$ and $\left(\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3\right)=(-3,1,5)$
$\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(-1,2,5)$ and $\left(\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3\right)=(-1,2,5)$
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 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|=2(5-10)-1(-15+5)+0(-6+1)=-10+10=0 \\
&
\end{aligned}
$
Given two lines are coplanar

$\begin{aligned}
& \left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
b_1 & b_2 & b_3 \\
d_1 & d_2 & d_3
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
x+3 & y-1 & z-5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|=0 \\
& (x+3)[5-10]-(y-1)[-15+5]+(z-5)[-6+1]=0 \\
& 5(\mathrm{x}+3)+10(\mathrm{y}-1)-5(\mathrm{z}-5)=0 \\
& (\div \text { by } 5) \Rightarrow(\mathrm{x}+3)-2(\mathrm{y}-1)+(\mathrm{z}-5)=0 \\
& \mathrm{x}+3-2 \mathrm{y}+2+\mathrm{z}-5=0 \\
& \mathrm{x}-2 \mathrm{y}+\mathrm{z}=0 \\
& \text { or } \\
& \left|\begin{array}{ccc}
x-x_2 & y-y_2 & z-z_2 \\
b_1 & b_2 & b_3 \\
d_1 & d_2 & d_3
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
x+1 & y-2 & z-5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|=0 \\
& (x+1)(5-10)-(y-2)(-15+5)+(z-5)(-6+1)=0 \\
& -5(\mathrm{x}+1)+10(\mathrm{y}-2)-5(\mathrm{z}-5)=0 \\
& (\mathrm{x}+1)-2(\mathrm{y}-2)+(\mathrm{z}-5)=0 \\
& \mathrm{x}+1-2 \mathrm{y}+4+\mathrm{z}-5=0 \\
& x-2 y+z=0 \\
&
\end{aligned}$