Exercise 6.5 - 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 $6.5$
Question 1.
Find the parametric form of vector equation and Cartesian equations of a straight line passing through (5,2, 8) and is perpendicular to the straight lines $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+s(2 \hat{i}-2 \hat{j}+\hat{k})$ and $\vec{r}=(2 \hat{i}-\hat{j}-3 \hat{k})+t(\hat{i}+2 \hat{j}+2 \hat{k})$
Solution:
Given point
$
\begin{aligned}
\vec{a} & =5 \vec{i}+2 \vec{j}+8 \vec{k}, \vec{b}=2 \vec{i}+\vec{j}-2 \vec{k}, \vec{d}=\vec{i}+2 \vec{j}+2 \vec{k} \\
\vec{b} \times \vec{d} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -2 & 1 \\
1 & 2 & 2
\end{array}\right|=-6 \vec{i}-3 \vec{j}+6 \vec{k}
\end{aligned}
$
$\therefore$ This' vector is perpendicular to both the given straight lines.
$\therefore$ The required straight line is $\vec{r}=\vec{a}+t(\vec{b} \times \vec{d})$
Cartesian equation:
$
\begin{aligned}
\vec{r} & =(5 \vec{i}+2 \vec{j}+8 \vec{k})+t(2 \vec{i}+\vec{j}-2 \vec{k}), t \in \mathbb{R} \\
\frac{x-5}{-6} & =\frac{y-2}{-3}=\frac{z-8}{6}(\mathrm{OR}) \\
\frac{x-5}{2} & =\frac{y-2}{1}=\frac{z-8}{-2}
\end{aligned}
$
Question 2.
Show that the lines $\vec{r}=(6 \hat{i}+\hat{j}+2 \hat{k})+s(\hat{i}+2 \hat{j}-3 \hat{k})$ and $\vec{r}=(3 \hat{i}+2 \hat{j}-2 \hat{k})+t(2 \hat{i}+4 \hat{j}-5 \hat{k}) \quad$ are skew lines and hence find the shortest distance between them.

Solution:
Given
$
\begin{aligned}
\vec{a} & =6 \vec{i}+\vec{j}+2 \vec{k}, \vec{b}=\vec{i}+2 \vec{j}-3 \vec{k} \\
\vec{c} & =3 \vec{i}+2 \vec{j}-2 \vec{k}, \vec{d}=2 \vec{i}+4 \vec{j}-5 \vec{k} \\
\vec{c}-\vec{a} & =-3 \vec{i}+\vec{j}-4 \vec{k} \\
{[(\vec{c}-\vec{a}) \vec{b} \vec{d}] } & =\left|\begin{array}{ccc}
-3 & 1 & -4 \\
1 & 2 & -3 \\
2 & 4 & -5
\end{array}\right|=-3(-10+12)-1(-5+6)-4(4-4) \\
& =-6-1 \Rightarrow-7 \neq 0
\end{aligned}
$
Given lines are skew lines
$
\begin{aligned}
\vec{b} \times \vec{d} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & 2 & -3 \\
2 & 4 & -5
\end{array}\right|=\vec{i}(-10+12)-\vec{j}(-5+6)+\vec{k}(4-4) \\
\vec{b} \times \vec{d} & =2 \vec{i}-\vec{j} \\
|\vec{b} \times \vec{d}| & =\sqrt{4+1}=\sqrt{5}
\end{aligned}
$
Shortest distance between skew lines $=\frac{|[(\vec{c}-\vec{a}) \vec{b} \vec{d}]|}{\sqrt{5}}=\frac{7}{\sqrt{5}}$ units.
Question 3.
If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-m}{2}=z$ intersect at a point, find the value of $m$.

Solution:

$
\begin{aligned}
& \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=s \text { (say) } \\
& (2 s+1,3 s-1,4 s+1) \\
& \frac{x-3}{1}=\frac{y-m}{2}=z=t \text { (say) } \\
& (t+3,2 t+m, t)
\end{aligned}
$
Given two lines are intersecting lines. So, equate the corresponding co-ordinates.
$
\begin{aligned}
& 2 s+1=t+3 \\
& 2 s-t=2
\end{aligned} \text {...(1) } \quad \mid 4 s+1=t \quad \ldots \text {.. (2) } \quad \mid 3 s-1=2 t+m
$

Substitute (2) in (1)
$
\begin{aligned}
& 2 s-4 s-1=2 \\
& -2 s=2+1 \quad \Rightarrow s=\frac{-3}{2} \\
& \text { sub } s=-\frac{3}{2} \text { in (1) } \\
&
\end{aligned}
$
(1) $\Rightarrow \quad 2\left(-\frac{3}{2}\right)-t=2$
$
-t=2+3 \quad \Rightarrow t=-5
$
Substitute $s=\frac{-3}{2}$ and $t=-5$ in (3)
$
\begin{aligned}
& \text { (3) } \Rightarrow \quad 3\left(-\frac{3}{2}\right)-1=2(-5)+m \\
& \frac{-9}{2}-1=-10+m \quad \Rightarrow \frac{-9}{2}+9=m \\
& \frac{9}{2}=m \\
&
\end{aligned}
$
Question 4.
Show that the lines $\frac{x-3}{3}=\frac{y-3}{-1}, z-1=0$ and $\frac{x-6}{2}=\frac{z-1}{3}, y-2=0$ intersect. Also find the point of intersection
Solution:

$
\begin{aligned}
\frac{x-3}{3} & =\frac{y-3}{-1}, z-1=0 \quad \Rightarrow z=1 \\
\frac{x-6}{2} & =\frac{z-1}{3}, y-2=0 \quad \Rightarrow y=2 \\
\left(x_1, y_1, z_1\right) & =(3,3,1) \text { and }\left(x_2, y_2, z_2\right)=(6,2,1) \\
\left(b_1, b_2, b_3\right) & =(3,-1,0) \text { and }\left(d_1, d_2, d_3\right)=(2,0,3)
\end{aligned}
$
Condition for intersection of two lines
$
\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}
3 & -1 & 0 \\
3 & -1 & 0 \\
2 & 0 & 3
\end{array}\right|=0 \quad \text { Since }\left(\mathrm{R}_1 \equiv \mathrm{R}_2\right)
\end{aligned}
$
$\therefore$ Given two lines are intersecting lines.
Any point on the first line
$
\frac{x-3}{3}=\frac{y-3}{-1}=\lambda \text { and } z=1
$
$
(3 \lambda+3,-\lambda+3,1)
$
Any point on the Second line
$
\begin{aligned}
& \frac{x-6}{2}=\frac{z-1}{3}=\mu \text { and } y=2 \\
& (2 \mu+6,2,3 \mu+1) \\
&
\end{aligned}
$
$\therefore$ The required point of intersection is $(6,2,1)$
Question 5.
Show that the straight lines $x+1=2 y=-12 z$ and $x=y+2=6 z-6$ are skew and hence find the shortest distance between them.
Solution:

$
\begin{aligned}
& x+1=2 y=-12 z \\
& x=y+2=6 z-6 \\
& x+1=\frac{y}{1 / 2}=\frac{z}{-1 / 12} \quad x=y+2=\frac{z-1}{1 / 6} \\
& \left(x_1, y_1, z_1\right)=(-1,0,0) \text { and }\left(x_2, y_2, z_2\right)=(0,-2,1) \\
& \left(b_1, b_2, b_3\right)=(1,1 / 2,-1 / 12) \text { and }\left(d_1, d_2, d_3\right)=(1,1,1 / 6) \\
&
\end{aligned}
$
Condition for skew lines
$
\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| \neq 0(\mathrm{OR})[(\vec{c}-\vec{a}) \vec{b} \vec{d}] \neq 0 \\
& {[(\vec{c}-\vec{a}) \vec{b} \vec{d}] }=\left|\begin{array}{ccc}
1 & -2 & 1 \\
1 & \frac{1}{2} & \frac{-1}{12} \\
1 & 1 & \frac{1}{6}
\end{array}\right|=1\left[\frac{1}{12}+\frac{1}{12}\right]+2\left[\frac{1}{6}+\frac{1}{12}\right]+1\left[1-\frac{1}{2}\right] \\
&=\frac{1}{6}+\frac{2}{6}+\frac{2}{12}+1-\frac{1}{2}=\frac{1}{6}+\frac{2}{6}+\frac{1}{6}+1-\frac{1}{2}=\frac{2}{3}+\frac{1}{2}=\frac{7}{6} \neq 0
\end{aligned}
$
Given Lines are skew lines.
$
\begin{aligned}
& \vec{b} \times \vec{d}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & \frac{1}{2} & -\frac{1}{12} \\
1 & 1 & \frac{1}{6}
\end{array}\right|=\vec{i}\left(\frac{1}{6}\right)-\vec{j}\left(\frac{3}{12}\right)+\vec{k}\left(\frac{1}{2}\right)=\frac{1}{6} \vec{i}-\frac{1}{4} \vec{j}+\frac{1}{2} \vec{k} \\
& |\vec{b} \times \vec{d}|=\sqrt{\frac{1}{36}+\frac{1}{16}+\frac{1}{4}}=\sqrt{\frac{16+36+144}{576}}=\sqrt{\frac{196}{576}}=\frac{14}{24}=\frac{7}{12}
\end{aligned}
$
Shortest distance between skew lines $=\frac{|[\vec{c}-\vec{a} \vec{b} \vec{d}]|}{|\vec{b} \times \vec{d}|}=\frac{7 / 6}{7 / 12}=\frac{7}{6} \times \frac{12}{7}=2$ units.

Question 6.
Find the parametric form of vector equation of the straight line passing through $(-1,2,1)$ and parallel to the straight line $\vec{r}=(2 \hat{i}+3 \hat{j}-\hat{k})+t(\hat{i}-2 \hat{j}+\hat{k})$ and hence find the shortest distance between the lines.
Solution:
Given point $\vec{a}=-\vec{i}+2 \vec{j}+\vec{k}, \vec{b}=\vec{i}-2 \vec{j}+\vec{k}$
$
\begin{aligned}
\vec{r} & =\vec{a}+t \vec{b} \\
\vec{r} & =(-\vec{i}+2 \vec{j}+\vec{k})+t(\vec{i}-2 \vec{j}+\vec{k})
\end{aligned}
$
Parallel to the straight line
$
\begin{aligned}
\vec{r} & =(2 \vec{i}+3 \vec{j}-\vec{k})+t(\vec{i}-2 \vec{j}+\vec{k}) \\
\text { Here } \vec{c} & =2 \vec{i}+3 \vec{j}-\vec{k} \\
\vec{c}-\vec{a} & =3 \vec{i}+\vec{j}-2 \vec{k} \\
(\vec{c}-\vec{a}) \times \vec{b} & =\left|\begin{array}{ccc}
i & \hat{j} & \hat{k} \\
3 & 1 & -2 \\
1 & -2 & 1
\end{array}\right|=-3 \vec{i}-5 \vec{j}-7 \vec{k} \\
|(\vec{c}-\vec{a}) \times \vec{b}| & =\sqrt{9+25+49}=\sqrt{83} \\
|\vec{b}| & =\sqrt{1+4+1}=\sqrt{6}
\end{aligned}
$
Shortest distance between parallel lines $=\frac{|(\vec{c}-\vec{a}) \times \vec{b}|}{|\vec{b}|}=\frac{\sqrt{83}}{\sqrt{6}}$ units.

Question 7.
Find the foot of the perpendicular drawn from the point $(5,4,2)$ to the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$. Also, find the equation of the perpendicular.
Solution:

Compare the given equation $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$ in to $\frac{x-x_1}{b_1}=\frac{y-y_1}{b_2}=\frac{z-z_1}{b_3}$ Here $\vec{b}=b_1 \vec{i}+b_2 \vec{j}+b_3 \vec{k}$ So, $\vec{b}=2 \hat{i}+3 \hat{j}-\vec{k}$
Any point on the line
$
\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=t \text { (say) }
$
The point $\mathrm{F}(2 t-1,3 t+3,-t+1)$

$
\begin{aligned}
& \overrightarrow{\mathrm{DF}}=\overrightarrow{\mathrm{OF}}-\overrightarrow{\mathrm{OD}} \\
& \overrightarrow{\mathrm{DF}}=(2 t-6) \vec{i}+(3 t-1) \vec{j}+(-t-1) \vec{k}
\end{aligned}
$
Since $\vec{b}$ is perpendicular to $\overrightarrow{\mathrm{DF}}$, we have
$
\begin{aligned}
& \vec{b} \cdot \overrightarrow{\mathrm{DF}}=0 \\
& 2(2 t-6)+3(3 t-1)-1(-t-1)=0 \\
& 4 t-12+9 t-3+t+1=0 \\
& 14 t-14=0 \\
& t=1
\end{aligned}
$
The co-ordinate of $\mathrm{F}$ is $(1,6,0)$. The equation of the perpendicular is
$
\begin{aligned}
\frac{x-x_1}{x_2-x_1} & =\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} \text { and } \mathrm{D}(5,4,2), \mathrm{F}(1,6,0) \\
\frac{x-5}{-4} & =\frac{y-4}{2}=\frac{z-2}{-2}
\end{aligned}
$