Exercise 6.8-Additional Problems - Chapter 6 - Applications of Vector Algebra - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Exercise 6.8-Additional Problems - Chapter 6 - Applications of Vector Algebra - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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

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