Exercise 6.5-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.
Find the shortest distance between the parallel line
$
\vec{r}=(\vec{i}-\vec{j})+t(2 \vec{i}-\vec{j}+\vec{k}) \text { and } \vec{r}=(2 \vec{i}+\vec{j}+\vec{k})+s(2 \vec{i}-\vec{j}+\vec{k})
$
Solution:
Compare the given equations with $\vec{r}=\vec{a}_1+t \vec{u}$ and $\vec{r}=\vec{a}_2+s \vec{u}$,
$
\begin{aligned}
& \vec{a}_1=\vec{i}-\vec{j} ; \vec{a}_2=2 \vec{i}+\vec{j}+\vec{k} \text { and } \vec{u}=2 \vec{i}-\vec{j}+\vec{k} \\
& \vec{a}_2-\vec{a}_1=\vec{i}+2 \vec{j}+\vec{k} \\
& \vec{u} \times\left(\vec{a}_2-\vec{a}_1\right)=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & -1 & 1 \\
1 & 2 & 1
\end{array}\right|=-3 \vec{i}-\vec{j}+5 \vec{k} \\
&\left|\vec{u} \times\left(\vec{a}_2-\vec{a}_1\right)\right|=\sqrt{9+1+25}=\sqrt{35} \\
&|\vec{u}|=\sqrt{4+1+1}=\sqrt{6}
\end{aligned}
$
The Distance between the parallel lines $=\frac{\left|\vec{u} \times\left(\vec{a}_2-\vec{a}_1\right)\right|}{|\vec{u}|}=\sqrt{\frac{35}{\sqrt{6}}}$

Question 2.
Show that the two lines $\vec{r}=(\vec{i}-\vec{j})+t(2 \vec{i}+\vec{k})$ and $\vec{r}=(2 \vec{i}-\vec{j})+s(\vec{i}+\vec{j}-\vec{k})$ skew lines and find the distance between them.
Solution:
Compare the given equations with $\vec{r}=\vec{a}_1+t \vec{u}$ and $\vec{r}=\vec{a}_2+s \vec{v}$
$\vec{a}_1=\vec{i}-\vec{j} ; \vec{a}_2=2 \vec{i}-\vec{j}$ and $\vec{u}=2 \vec{i}+\vec{k}, \vec{v}=\vec{i}+\vec{j}-\vec{k}$
$
\begin{aligned}
\vec{a}_2-\vec{a}_1 & =\vec{i} \\
{\left[\left(\vec{a}_2-\vec{a}_1\right) \vec{u} \vec{v}\right] } & =\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 & 0 & 1 \\
1 & 1 & -1
\end{array}\right|=-1 \neq 0
\end{aligned}
$
$\therefore$ They are skew lines.
$
\begin{aligned}
& \vec{u} \times \vec{v}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 0 & 1 \\
1 & 1 & -1
\end{array}\right|=-\vec{i}+3 \vec{j}+2 \vec{k} \\
& |\vec{u} \times \vec{v}|=\sqrt{14}
\end{aligned}
$
Shortest distance between the lines $=\left[\frac{\left(\vec{a}_2-\vec{a}_1\right) \vec{u} \vec{v}}{|\vec{u} \times \vec{v}|}\right]$
From (1) shortest distance between them is $\frac{1}{\sqrt{14}}$

Question 3.
Show that the lines $\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $\frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3}$ intersect and hence find the point of intersection

Solution:
The condition for intersecting is $\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=0$
Compare with $\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$, we get
$
\begin{aligned}
& \left(x_1, y_1, z_1\right)=(1,1,-1) ;\left(x_2, y_2, z_2\right)=(4,0,-1) \\
& \left(l_1, m_1, n_1\right)=(3,-1,0) ;\left(l_2, m_2, n_2\right)=(2,0,3)
\end{aligned}
$
The determinant becomes
$
\left|\begin{array}{ccc}
3 & -1 & 0 \\
3 & -1 & 0 \\
2 & 0 & 3
\end{array}\right|=0
$
Note that $\vec{u}$ and $\vec{v}$ are not parallel.
$\therefore$ The lines are intersecting lines.
Point of intersection:
Take
$
\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda
$
$\therefore$ Any point on the line is of the form $(3 \lambda+1,-\lambda+1,-1)$
Take
$
\frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3}=\mu
$
Any point on this line is of the form $(2 \mu+4,0,3 \mu-1)$ Since they are intersecting, for some $\lambda, \mu$ $(3 \lambda+1,-\lambda+1,-1)=(2 \mu+4,0,3 \mu-1) \Rightarrow \lambda=1$ and $\mu=0$
To find the point of intersection either take $\lambda=1$ or $\mu=0$
$\therefore$ The point of intersection is $(4,0,-1)$,

Question 4.
Find the shortest distance between the skew lines.
$
\vec{r}=(\vec{i}-\vec{j})+\lambda(2 \vec{i}+\vec{j}+\vec{k}) \text { and } \vec{r}=(\vec{i}+\vec{j}-\vec{k})+\mu(2 \vec{i}-\vec{j}-\vec{k})
$
Solution:
Compare the given equation with $\vec{r}=\vec{a}_1+t \vec{u}$ and $\vec{r}=\vec{a}_2+s \vec{v}$
$
\begin{aligned}
& \vec{a}_1=\vec{i}-\vec{j} ; \vec{a}_2=\vec{i}+\vec{j}-\vec{k} ; \vec{u}=2 \vec{i}+\vec{j}+\vec{k} ; \vec{v}=2 \vec{i}-\vec{j}-\vec{k} \\
& \vec{a}_2-\vec{a}_1=2 \vec{j}-\vec{k} \text { and } \vec{u} \times \vec{v}=4 \vec{j}-4 \vec{k} \\
& {\left[\left(\vec{a}_2-\vec{a}_1\right) \vec{u} \vec{v}\right] }=\left|\begin{array}{ccc}
0 & 2 & -1 \\
2 & 1 & 1 \\
2 & -1 & 1
\end{array}\right|=12 \\
&|\vec{u} \times \vec{v}|=4 \sqrt{2} \\
& \text { distance }=\frac{\left|\left[\left(\vec{a}_2-\vec{a} 1\right) \vec{u} \vec{v}\right]\right|}{|\vec{u} \vec{v}|}=\frac{12}{4 \sqrt{2}}=\frac{3}{\sqrt{2}}
\end{aligned}
$

Question 5.
Find the shortest distance between the parallel lines.
$
\vec{r}=(2 \vec{i}-\vec{j}-\vec{k})+t(\vec{i}-2 \vec{j}+3 \vec{k}) \text { and } \vec{r}=(\vec{i}-2 \vec{j}+\vec{k})+s(\vec{i}-2 \vec{j}+3 \vec{k})
$
Solution:
Comparing the given equations with $\vec{r}=\vec{a}_1+t \vec{u}$ and $\vec{r}=\vec{a}_2+s \vec{u}$ we get,
$
\begin{aligned}
\vec{a}_1 & =2 \vec{i}-\vec{j}-\vec{k} \\
\vec{a}_2 & =\vec{i}-2 \vec{j}+\vec{k} \text { and } \vec{u}=\vec{i}-2 \vec{j}+3 \vec{k} \\
\vec{a}_2-\vec{a}_1 & =(\vec{i}-2 \vec{j}+\vec{k})-(2 \vec{i}-\vec{j}-\vec{k})=-\vec{i}-\vec{j}+2 \vec{k} \\
\vec{u} \times\left(\vec{a}_2-\vec{a}_1\right) & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & -2 & 3 \\
-1 & -1 & 2
\end{array}\right|=\vec{i}[-4+3]-\vec{j}[2+3]+\vec{k}[-1-2]=-\vec{i}-5 \vec{j}-3 \vec{k} \\
\left|\vec{u} \times\left(\vec{a}_2-\vec{a}_1\right)\right| & =\sqrt{1+25+9}=\sqrt{35} \\
|\vec{u}| & =\sqrt{1+4+9}=\sqrt{14}
\end{aligned}
$
The distance between the parallel lines $=\frac{\left|\vec{u} \times\left(\vec{a}_2-\vec{a}_1\right)\right|}{|\vec{u}|}=\frac{\sqrt{35}}{\sqrt{14}}=\sqrt{\frac{35}{14}}=\sqrt{\frac{5}{2}}$ units.