Question 13 Marks
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2 \hat{i}-\hat{j}+4 \hat{k}$ and is in the direction $\hat{i}+2 \hat{j}-\hat{k}$
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Question 23 Marks
Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are
$
\begin{aligned}
\vec{r} & =\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})
\end{aligned}
$
$
\begin{aligned}
\vec{r} & =\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})
\end{aligned}
$
Answer
View full question & answer→Given lines
$
\begin{aligned}
\vec{r} & =\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})
\end{aligned}
$
Comparing equations (1) and (2) with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and
$
\begin{aligned}
\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2} & \\
& \overrightarrow{a_1}=\hat{i}+\hat{j}, \overrightarrow{b_1}=2 \hat{i}-\hat{j}+\hat{k} \\
& \overrightarrow{a_2}=2 \hat{i}+\hat{j}-\hat{k} \text { and } \overrightarrow{b_2}=3 \hat{i}-5 \hat{j}+2 \hat{k}
\end{aligned}
$
So, $\quad \overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{k}$
and
$
\begin{aligned}
\overrightarrow{b_1} \times \overrightarrow{b_2} & =(2 \hat{i}-\hat{j}+\hat{k}) \times(3 \hat{i}-5 \hat{j}+2 \hat{k}) \\
& =\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|
\end{aligned}
$
$
\begin{array}{l}
=(-2+5) \hat{i}-(4-3) \hat{j}+(-10+3) \hat{k} \\
=3 \hat{i}-\hat{j}-\hat{k}
\end{array}
$
Thus $\quad\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{9+1+49}=\sqrt{59}$
Therefore the shortest distance $
d=\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2} \times \overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right|
$
Putting the values
$
\begin{aligned}
d & =\left|\frac{(3 \hat{i}-\hat{j}-7 \hat{k}) \cdot(\hat{i}-\hat{k})}{\sqrt{59}}\right| \\
& =\left|\frac{3-0+7}{\sqrt{59}}\right|=\frac{10}{\sqrt{59}}
\end{aligned}
$
$
\begin{aligned}
\vec{r} & =\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})
\end{aligned}
$
Comparing equations (1) and (2) with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and
$
\begin{aligned}
\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2} & \\
& \overrightarrow{a_1}=\hat{i}+\hat{j}, \overrightarrow{b_1}=2 \hat{i}-\hat{j}+\hat{k} \\
& \overrightarrow{a_2}=2 \hat{i}+\hat{j}-\hat{k} \text { and } \overrightarrow{b_2}=3 \hat{i}-5 \hat{j}+2 \hat{k}
\end{aligned}
$
So, $\quad \overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{k}$
and
$
\begin{aligned}
\overrightarrow{b_1} \times \overrightarrow{b_2} & =(2 \hat{i}-\hat{j}+\hat{k}) \times(3 \hat{i}-5 \hat{j}+2 \hat{k}) \\
& =\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|
\end{aligned}
$
$
\begin{array}{l}
=(-2+5) \hat{i}-(4-3) \hat{j}+(-10+3) \hat{k} \\
=3 \hat{i}-\hat{j}-\hat{k}
\end{array}
$
Thus $\quad\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{9+1+49}=\sqrt{59}$
Therefore the shortest distance $
d=\left|\frac{\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \cdot\left(\overrightarrow{a_2} \times \overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right|
$
Putting the values
$
\begin{aligned}
d & =\left|\frac{(3 \hat{i}-\hat{j}-7 \hat{k}) \cdot(\hat{i}-\hat{k})}{\sqrt{59}}\right| \\
& =\left|\frac{3-0+7}{\sqrt{59}}\right|=\frac{10}{\sqrt{59}}
\end{aligned}
$
Question 33 Marks
Find the shortest distance between the following given lines $l_1$ and $l_2$
$
\begin{array}{l}
\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\
\text { and } \vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{array}
$
$
\begin{array}{l}
\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\
\text { and } \vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{array}
$
Answer
View full question & answer→Both the lines are parallel, because
$\overrightarrow{a_1}=\hat{i}+2 \hat{j}-4 \hat{k}, \quad \overrightarrow{a_2}=3 \hat{i}+3 \hat{j}-5 \hat{k}$ and
$\overrightarrow{b_1}=2 \hat{i}+3 \hat{j}+6 \hat{k}=\vec{b}_2$
Hence the distance between the parallel lines
$
\begin{array}{l}
d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right| \\
\vec{a}_2-\vec{a}_1=(3 \hat{i}+3 \hat{j}-5 \hat{k})-(\hat{i}+2 \hat{j}-4 \hat{k})=2 \hat{i}+\hat{j}-\hat{k} \\
\therefore \vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 6 \\
2 & 1 & -1
\end{array}\right|=(-3-6) \hat{i} \\
-(-2-12) \hat{j}+(2-6) \hat{k} \\
=-9 \hat{i}+14 \hat{j}-4 \hat{k} \\
|\vec{b}|=\sqrt{(2)^2+(3)^2+(6)^2} \\
=\sqrt{4+9+36}=\sqrt{49}=7 \\
\text { Hence }\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right|=\frac{\sqrt{(-9)^2+(14)^2+(-4)^2}}{7} \\
=\frac{\sqrt{81+196+16}}{7}=\frac{\sqrt{293}}{7} \text { }
\end{array}
$
$\overrightarrow{a_1}=\hat{i}+2 \hat{j}-4 \hat{k}, \quad \overrightarrow{a_2}=3 \hat{i}+3 \hat{j}-5 \hat{k}$ and
$\overrightarrow{b_1}=2 \hat{i}+3 \hat{j}+6 \hat{k}=\vec{b}_2$
Hence the distance between the parallel lines
$
\begin{array}{l}
d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right| \\
\vec{a}_2-\vec{a}_1=(3 \hat{i}+3 \hat{j}-5 \hat{k})-(\hat{i}+2 \hat{j}-4 \hat{k})=2 \hat{i}+\hat{j}-\hat{k} \\
\therefore \vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 6 \\
2 & 1 & -1
\end{array}\right|=(-3-6) \hat{i} \\
-(-2-12) \hat{j}+(2-6) \hat{k} \\
=-9 \hat{i}+14 \hat{j}-4 \hat{k} \\
|\vec{b}|=\sqrt{(2)^2+(3)^2+(6)^2} \\
=\sqrt{4+9+36}=\sqrt{49}=7 \\
\text { Hence }\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right|=\frac{\sqrt{(-9)^2+(14)^2+(-4)^2}}{7} \\
=\frac{\sqrt{81+196+16}}{7}=\frac{\sqrt{293}}{7} \text { }
\end{array}
$
Question 43 Marks
If the direction cosines of any variable line in its two adjacent position are $l , m , n$ and $l +\delta, m +\delta m$, $n +\delta n$ the angle between those position be $\delta \theta$, then prove that
$
(\delta \theta)^2=(\delta l)^2+(\delta m)^2+(\delta n)^2
$
View full question & answer→$
(\delta \theta)^2=(\delta l)^2+(\delta m)^2+(\delta n)^2
$
Question 53 Marks
Find the shortest distance between the following lines:
$
\begin{array}{l}
\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\
\vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+\mu(4 \hat{i}+6 \hat{j}+8 \hat{k})
\end{array}
$
$
\begin{array}{l}
\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\
\vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+\mu(4 \hat{i}+6 \hat{j}+8 \hat{k})
\end{array}
$
Answer
View full question & answer→Comparing the equations of the given line with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\lambda \overrightarrow{b_2}$, we get
$
\begin{array}{l}
\overrightarrow{a_1}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{b_1}=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
\overrightarrow{a_2}=2 \hat{i}+4 \hat{j}+5 \hat{k} \text { and } \overrightarrow{b_2}=4 \hat{i}+6 \hat{j}+8 \hat{k} \\
\quad \therefore \overrightarrow{b_2}=\vec{b}
\end{array}
$
So $\overrightarrow{a_2}-\vec{a}_1=2 \hat{i}+4 \hat{j}+5 \hat{k}-\hat{i}-2 \hat{j}-3 \hat{k}=\hat{i}+2 \hat{j}+2 \hat{k}$ Here both the lines are parallel to each other. Hence the distance between the parallel lines$
d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right|
$
Now finding the value of $\vec{b} \times\left(a_2-a_1\right)$,
$
\begin{aligned}
\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
1 & 2 & 2
\end{array}\right| & =(6-8) \hat{i}-(4-4) \hat{j}+(4-3) \hat{k} \\
& =-2 \hat{i}+\hat{k} \\
\therefore\left|\vec{b} \times\left(\vec{a}_2-\overrightarrow{a_1}\right)\right| & =\sqrt{(-2)^2+(1)^2}=\sqrt{5} \\
|\vec{b}| & =\sqrt{(2)^3+(3)^2+(4)^2}=\sqrt{4+9+16} \\
& =\sqrt{29}
\end{aligned}
$
Hence the distance between the parallel lines.
$
\begin{array}{l}
d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right| \\
d=\frac{\sqrt{5}}{\sqrt{29}} \\
d=\frac{\sqrt{5}}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}}=\frac{\sqrt{145}}{29}
\end{array}
$
Ans.
$
\begin{array}{l}
\overrightarrow{a_1}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{b_1}=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
\overrightarrow{a_2}=2 \hat{i}+4 \hat{j}+5 \hat{k} \text { and } \overrightarrow{b_2}=4 \hat{i}+6 \hat{j}+8 \hat{k} \\
\quad \therefore \overrightarrow{b_2}=\vec{b}
\end{array}
$
So $\overrightarrow{a_2}-\vec{a}_1=2 \hat{i}+4 \hat{j}+5 \hat{k}-\hat{i}-2 \hat{j}-3 \hat{k}=\hat{i}+2 \hat{j}+2 \hat{k}$ Here both the lines are parallel to each other. Hence the distance between the parallel lines$
d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right|
$
Now finding the value of $\vec{b} \times\left(a_2-a_1\right)$,
$
\begin{aligned}
\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
1 & 2 & 2
\end{array}\right| & =(6-8) \hat{i}-(4-4) \hat{j}+(4-3) \hat{k} \\
& =-2 \hat{i}+\hat{k} \\
\therefore\left|\vec{b} \times\left(\vec{a}_2-\overrightarrow{a_1}\right)\right| & =\sqrt{(-2)^2+(1)^2}=\sqrt{5} \\
|\vec{b}| & =\sqrt{(2)^3+(3)^2+(4)^2}=\sqrt{4+9+16} \\
& =\sqrt{29}
\end{aligned}
$
Hence the distance between the parallel lines.
$
\begin{array}{l}
d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right| \\
d=\frac{\sqrt{5}}{\sqrt{29}} \\
d=\frac{\sqrt{5}}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}}=\frac{\sqrt{145}}{29}
\end{array}
$
Ans.
Question 63 Marks
Find the coordinates of the foot of the perpendicular and its length drawn from the point $P(5,4,2)$ to the line
$
\vec{r}=-\hat{i}+3 \hat{j}+\hat{k}+\lambda \quad(2 \hat{i}+3 \hat{j}-\hat{k})
$
$
\vec{r}=-\hat{i}+3 \hat{j}+\hat{k}+\lambda \quad(2 \hat{i}+3 \hat{j}-\hat{k})
$
Answer
View full question & answer→$
\vec{r}=-\hat{i}+3 \hat{j}+\hat{k}+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})
$
Equation of the line in Cartesian form are
$
\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=\lambda
$
\vec{r}=-\hat{i}+3 \hat{j}+\hat{k}+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})
$
Equation of the line in Cartesian form are
$
\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=\lambda
$
Question 73 Marks
Find the length of the perpendicular and coordinates of the foot of the perpendicular drawn from point $P(2,-1,5)$ on the given line $\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}$.
Answer
View full question & answer→The equations of the given line are
$
\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=\lambda \text { (say) }
$
$
\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=\lambda \text { (say) }
$
Question 83 Marks
Find the angle between the lines
$
\begin{array}{ll}
& 2 x=3 y=-z \\
\text { and } & 6 x=-y=-4 z .
\end{array}
$
View full question & answer→$
\begin{array}{ll}
& 2 x=3 y=-z \\
\text { and } & 6 x=-y=-4 z .
\end{array}
$
Question 93 Marks
From point $P(1,2,3)$ perpendicular $P N$ is drawn to the line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$. Then find the following :
(i) Coordinates of point N
(ii) Length of PN
(i) Coordinates of point N
(ii) Length of PN
Answer
View full question & answer→From point $P (1,2,3)$ perpendicular PN is drawn to the line
$
\frac{x-2}{3}=\frac{y-2}{4}=\frac{z-4}{5}=r \text { (say) }
$
Then the coordinates of N can be taken as $(3 r+2,4 r$ $+3,5 r+4$ ).
Direction ratios of PN are
$
\begin{array}{r}
3 r+2-1,4 r+3-2,5 r+4-3 \\
3 r+1,4 r+1,5 r+1
\end{array}
$
i.e., Since PN is perpendicular to the line, therefore,
$
\frac{x-2}{3}=\frac{y-2}{4}=\frac{z-4}{5}=r \text { (say) }
$
Then the coordinates of N can be taken as $(3 r+2,4 r$ $+3,5 r+4$ ).
Direction ratios of PN are
$
\begin{array}{r}
3 r+2-1,4 r+3-2,5 r+4-3 \\
3 r+1,4 r+1,5 r+1
\end{array}
$
i.e., Since PN is perpendicular to the line, therefore,