3 Marks

Question 13 Marks

Find the length of the perpendicular drow from the point (5, 4, -1) to the line $\vec{\text{r}}=\hat{\text{i}}+\lambda\big(2\hat{\text{i}}+9\hat{\text{j}}+5\hat{\text{k}}\big).$

Answer

Let the point (5, 4, -1) be P and the point through which the line passes be Q(1, 0, 0).

The line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+9\hat{\text{k}}+5\hat{\text{k}}.$

Now,

$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&9&5\\-4&-4&1\end{vmatrix}$

$=29\hat{\text{i}}-22\hat{\text{j}}+28\hat{\text{k}}$

$\Rightarrow\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{29^2+(-22)^2+28^2}$

$=\sqrt{841+484+784}$

$=\sqrt{2109}$

$\big|\vec{\text{b}}\big|=\sqrt{2^+9^2+5^2}$

$=\sqrt{4+81+25}$

$=\sqrt{110}$

$\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$

$=\frac{\sqrt{2109}}{\sqrt{110}}$

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Question 23 Marks

Write the direction cosines of the line $\frac{\text{x}-2}{2}=\frac{2\text{y}-5}{-3},\text{z}=2.$

Answer

We have

$\frac{\text{x}-2}{2}=\frac{2\text{y}-5}{-3},\text{z}=2$

The equation of given line can be re-written as

$\frac{\text{x}-2}{2}=\frac{\text{y}-\frac{5}{2}}{-\frac{3}{2}}=\frac{\text{z}-2}{0}$

$\frac{\text{x}-2}{4}=\frac{\text{y}-\frac{5}{2}}{-3}=\frac{\text{z}-2}{0}$

The direction ratios of the given line are proportional to 4, -3, 0.

Hence, the direction cosines of the given line are proportional to

$\frac{4}{\sqrt{4^2+(-3)^2+0^2}},\frac{-3}{\sqrt{4^2+(-3)^2+0^2}},\frac{0}{\sqrt{4^2+(-3)^2+0^2}}$

$=\frac{4}{5},\frac{-3}{5},0$

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Question 33 Marks

Cartesian equations of a line AB are $\frac{2\text{x}-1}{2}=\frac{4-\text{y}}{7}=\frac{\text{z}+1}{2}.$ Write the direction ratios of a parallel to AB.

Answer

$\frac{2\text{x}-1}{2}=\frac{4-\text{y}}{7}=\frac{\text{z}+1}{2}$

The equation of the line AB can be re-written as

$\frac{\text{x}-\frac{1}{2}}{1}=\frac{\text{y}-4}{-7}=\frac{\text{z}+1}{2}$

The direction ratios of the line parallel to AB are proportional to 1, -7, 2.

Also, the diraction cosines of the line parallel to AB are proportional to

$\frac{1}{\sqrt{1^2+(-7)^2+2^2}},\frac{-7}{\sqrt{1^2+(-7)^2+2^2}},\frac{2}{\sqrt{1^2+(-7)^2+2^2}}$

$=\frac{1}{\sqrt{54}},\frac{-7}{\sqrt{54}},\frac{2}{\sqrt{54}}$

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Question 43 Marks

Find the angle between the lines $2\text{x}=3\text{y}=-\text{z}$ and $6\text{x}=-\text{y}=-4\text{z}.$

Answer

The equations of the given lines can be re-writen as
$\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$ and $\frac{\text{x}}{2}=\frac{\text{y}}{-12}=\frac{\text{z}}{-3}$
We know that angle between the lines $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ is given by $\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}.$
Let $\theta$ be the angle between the given lines.
$\therefore\cos\theta=\frac{3\times2+2\times(-12)+(-6)\times(-3)}{\sqrt{3^2+2^2+(-6)^2}\sqrt{2^2+(-12)^2+(-3)^2}}$
$=\frac{6-24+18}{\sqrt{49}\sqrt{157}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$
Thus, the angle between the given lines is $\frac{\pi}{2}.$

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Question 53 Marks

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3, -6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer

The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (-4, 3, -6), and (2, 9, 2) respectively.
The direction ratios of AB are (4 - 1) = 3, (5 - 2) = 3, and (7 - 3) = 4
The direction ratios of CD are (2 - (-4)) = 6, (9 - 3) = 6, and (2 - (-6)) = 8
It can be seen that, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Therefore, AB is parallel to CD.
Thus, the angle between AB and CD is either 0° or 180°.

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Question 63 Marks

The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

Answer

$\text{x}=\text{ay+b},$
$\text{z}=\text{cy+d}$
$\frac{\text{x}-\text{b}}{\text{a}}=\frac{\text{y}}{1}=\frac{\text{z}-\text{d}}{\text{c}}=\lambda$ (say)
So DR's of line are (a, 1, c)
from above equation, we can write
$\text{x}=\text{a}\lambda+\text{b}$
$\text{y}=\lambda$
$\text{z}=\text{c}\lambda+\text{d}$
So vector equation of line is
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(\text{b}\hat{\text{i}}+\text{d}\hat{\text{k}})+\lambda(\text{a}\hat{\text{i}}+\text{x}\hat{\text{j}}+\text{c}\hat{\text{k}}\big)$

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Question 73 Marks

Find the equation of the line passing through the points (2, 1, 3) and perpendicular to the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}$ and $\frac{\text{x}}{-3}=\frac{\text{y}}{2}=\frac{\text{z}}{5}$

Answer

Let:
$\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
Since the required line is perpendicular to the lines parallel to the vectors
$\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}},$ it is parallel to the vector $\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2.$
Now,
$\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\-3&2&5\end{vmatrix}$
$=4\hat{\text{i}}-14\hat{\text{j}}+8\hat{\text{k}}$
$=2\big(2\hat{\text{i}}-7\hat{\text{j}}+4\hat{\text{k}}\big)$
Thus, the diraction ratios of the required line are proportional to 2, -7, 4.
The equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2, -7, 4 is $\frac{\text{x}-2}{2}=\frac{\text{y}-1}{-7}=\frac{\text{z}-3}{4}.$

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Question 83 Marks

Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (-1, -2, 1) and, (1, 2, 5).

Answer

The diraction ratios of a line passing through the points (4, 7, 8) and (2, 3, 4) are
(4 - 2, 7 - 3, 8 - 4)
= (2, 4, 4)
The direction ratios of a line passing through the points
(-1, -2, 1) and (1, 2, 5)are
(-1 -1, -2 -2, 1 - 5)
= (-2, -4, -4)
The direction ratios are proportional.
$\frac{2}{-2}=\frac{4}{-4}=\frac{4}{-4}$
Hence, the lines are mutually parallel.

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Question 93 Marks

Find the equation of the line passing through the points (1, 2, -4) and parallel to the line $\frac{\text{x}-3}{4}=\frac{\text{y}-5}{2}=\frac{\text{z}+1}{3}.$

Answer

The direction ratios of the line parallel to line $\frac{\text{x}-3}{4}=\frac{\text{y}-5}{2}=\frac{\text{z}+1}{3}$ are proportional to 4, 2, 3.
Equation of the required line passing through the point (1, 2, -4) having direction ratios proportional to 4, 2, 3 is
$\frac{\text{x}-1}{4}=\frac{\text{y}-2}{2}=\frac{\text{z}-(-4)}{3}$
$=\frac{\text{x}-1}{4}=\frac{\text{y}-2}{2}=\frac{\text{z}+4}{3}$

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Question 103 Marks

Find the distance of the point (2, 4, -1) from the line $\frac{\text{x}+5}{1}=\frac{\text{y}+3}{4}=\frac{\text{z}-6}{-9}.$

Answer

Let P = (2, 4, -1)
In order to find the distance we need to find a point Q on the line.
So, let take this point as required point.
Also line is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+4\hat{\text{j}}-9\hat{\text{k}}.$
Now, $\overrightarrow{\text{PQ}}=\big(-5\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}\big)-\big(2\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}}\big)=-7\hat{\text{i}}-7\hat{\text{j}}+7\hat{\text{k}}$
$\vec{\text{b}}\times\overrightarrow{\text{PQ}}=\begin{bmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&4&-9\\-7&-7&7 \end{bmatrix}=-35\hat{\text{i}}+56\hat{\text{j}}+21\hat{\text{k}}$
$\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{1225+3136+441}=\sqrt{4802}$
$\big|\vec{\text{b}}\big|=\sqrt{1+16+81}=\sqrt{98}$
$\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}=\frac{\sqrt{4802}}{\sqrt{98}}=7$

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Question 113 Marks

Show that the lines $\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ are perpendicular to each

Answer

we have
$\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$
These equations can be-written as
$\frac{\text{x}-5}{7}=\frac{\text{y}-(-2)}{-5}=\frac{\text{z}-0}{1}\dots(1)$
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{2}=\frac{\text{z}-0}{3}\dots(2)$
$\therefore\vec{\text{m}_1}=$ vector parallel to line (1) $=7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{m}}_2=$ vector parallel to line (2) $=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Now,
$\vec{\text{m}}_1\vec{\text{m}}_2=\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=7-10+3$
$=0$
Hence, the given two lines are perpendicular to each other.

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Question 123 Marks

The cartesian equation of a line AB are $\frac{2\text{x}-1}{\sqrt{3}}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{3}.$ Find the direction cosines of a line parallel to AB.

Answer

We have
$\frac{2\text{x}-1}{\sqrt{3}}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{3}$
The equation of the line AB can be re-written as
$\frac{\text{x}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{3}$
$=\frac{\text{x}-\frac{1}{2}}{\sqrt{3}}=\frac{\text{y}+2}{4}=\frac{\text{z}-3}{6}$
Thus, the direction ratios of the line parallel to AB are proportional to 3, 4, 6.
Hence, the direction cosines of the line parallel to AB are proportional to
$\frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2+4^2+6^2}},\frac{4}{\sqrt{(\sqrt{3})^2+4^2+6^2}},\frac{6}{\sqrt{(\sqrt{3})^2+4^2+6^2}}$
$=\frac{\sqrt{3}}{\sqrt{55}},\frac{4}{\sqrt{55}},\frac{6}{\sqrt{55}}$

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Question 133 Marks

A line passes throuth the point with position vector $2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and is in the direction of $3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}.$ Find equations of the line in vector and cartesian form.

Answer

We know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}.$
Here,
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
So, the vector equation of the required line is
$\vec{\text{r}}=\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.

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Question 143 Marks

Show that the lines $\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ are perpendicular to each other.

Answer

The direction ratios of the lines $\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ are proportional to 7, -5, 1 and 1, 2, 3, respectiveiy.
Let:
$\vec{\text{b}}_1=7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Now,
$\vec{\text{b}}_1.\vec{\text{b}}_2=\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=7-10+3$
$=0$
$\therefore\vec{\text{b}}_1\perp\vec{\text{b}}_2$
Hence, the given lines are perpendicular to each other.

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Question 153 Marks

Write the direction cosines of the line whose cartesian equations are 6x -2 = 3y + 1 =2z - 4.

Answer

We have
6x -2 = 3y + 1 =2z - 4
The equation of given line can be re-written as
$\frac{\text{x}-\frac{1}{3}}{\frac{1}{6}}=\frac{\text{y}+\frac{1}{3}}{\frac{1}{3}}=\frac{\text{z}-2}{\frac{1}{2}}$
$\Rightarrow\frac{\text{x}-\frac{1}{3}}{1}=\frac{\text{y}+\frac{1}{3}}{2}=\frac{\text{z}-2}{3}$
The direction ratios of the line parallel to AB are proportional to 1, 2, 3.
Hence, the direction cosines of the line parallel to AB are proportional to
$\frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}}$
$=\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$

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Question 163 Marks

Write the angle between the lines $\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}-2}{1}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}}{2}=\frac{\text{z}-1}{3}.$

Answer

We have
$\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}-2}{1}$
$\frac{\text{x}-1}{1}=\frac{\text{y}}{2}=\frac{\text{z}-1}{3}$
The given lines are parallel to the vectors $\vec{\text{b}}_1=7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)}{\sqrt{7^2+(-5)^2+1^2}\sqrt{1^2+2^2+3^2}}$
$=\frac{7-10+3}{\sqrt{49+25+1}\sqrt{1+4+9}}$
$=0$

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Question 173 Marks

Find the value of $\lambda$ so that the following lines are perpendicular to each other.

$\frac{\text{x}-5}{5\lambda+2}=\frac{2-\text{y}}{5}=\frac{1-\text{z}}{-1},\frac{\text{x}}{1}=\frac{2\text{y}+1}{4\lambda}=\frac{1-\text{z}}{-3}$

Answer

The equation of the given lines $\frac{\text{x}-5}{5\lambda+2}=\frac{2-\text{y}}{5}=\frac{1-\text{z}}{-1}$ and $\frac{\text{x}}{1}=\frac{2\text{y}+1}{4\lambda}=\frac{1-\text{z}}{-3}$ can be re-written as

$\frac{\text{x}-5}{5\lambda+2}=\frac{\text{y}-2}{-5}=\frac{\text{z}-1}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}+\frac{1}{2}}{2\lambda}=\frac{\text{z}-1}{3}$

Since the given lines are pependicular to each other, we have

$(5\lambda+2)1-5(2\lambda)+1(3)=0$

$\Rightarrow5\lambda=5$

$\Rightarrow\lambda=1$

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Question 183 Marks

Find in vector form as wel as in cartesian form, the equation of the line passing through the points A(1, 2, -1) and B(2, 1, 1).

Answer

We know that, equation of line passing through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is
$\frac{\text{x}-\text{x}_1}{\text{x}_2-\text{x}_1}=\frac{\text{y}-\text{y}_1}{\text{y}_2-\text{y}_1}=\frac{\text{z}-\text{z}_1}{\text{z}_2-\text{z}_1}\dots(1)$
Here, $\big(\text{x}_1,\text{y}_1,\text{z}_1\big)=\text{A}(1,2,-1)$
$\big(\text{x}_2,\text{y}_2,\text{z}_2\big)=\text{B}(2,1,1)$
Using equation (1), equation of line AB
$\frac{\text{x}-1}{2-1}=\frac{\text{y}-2}{1-2}=\frac{\text{z}+1}{1+1}$
$\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}+1}{2}=\lambda$ (Say)
$\text{x}=\lambda+1,\text{y}=-\lambda+2,\text{z}=2\lambda-1$
vector form of equation of line Ab is,
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(\lambda+1)\hat{\text{i}}+(-\lambda+2)\hat{\text{j}}+(2\lambda-1)\hat{\text{k}}$
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$

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Question 193 Marks

Find the points on the line $\frac{\text{x}+2}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-3}{2}$ at a distance of 5 units from the point P(1, 3, 3).

Answer

The coordinates of any point on the line $\frac{\text{x}+2}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-3}{2}$ are given by
$\frac{\text{x}+2}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-3}{2}=\lambda$
$\Rightarrow\text{x}=3\lambda-2,\text{y}=2\lambda-1,\text{z}=2\lambda+3\dots(1)$
Let the coordinates of the desired point be $(3\lambda-2,2\lambda-1,2\lambda+3)$
The distance between this point and (1, 3, 3) is 5 units.
$\therefore\sqrt{(3\lambda-2-1)^2+(2\lambda-1-3)^2+(2\lambda+3-3)^2}=5$
$\Rightarrow(3\lambda-3)^2+(2\lambda-4)^2+(2\lambda)^2=25$
$\Rightarrow17\lambda^2-34\lambda=0$
$\Rightarrow\lambda(\lambda-2)=0$
$\Rightarrow\lambda=0$ or $2$
Substituting the values of $\lambda$ in (1), we get the coordinates of the desired point as (-2, -1, 3) and (4, 3, 7).

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Question 203 Marks

Show that the line through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).

Answer

Suppose vector $\vec{\text{a}}$ is passing through the points (1, -1, 2) and (3, 4, -2) and $\vec{\text{b}}$ passing through the points (0, 3, 2) and (3, 5, 6).
Then,
$\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
Now,
$\vec{\text{a}}.\vec{\text{b}}=\big(2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}\big).\big(3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)=0$
Hence, the given lines are perpendicular to each other.

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Question 213 Marks

It the lines $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{2\lambda}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}-6}{-5}$ are perpendicular, find the value of $\lambda.$

Answer

The diraction of ratios of the lines, $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{2\lambda}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}-6}{-5},$ are -3, 2k, 2 and 3k, 1, -5 respectiveiy.
It is know that two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂, are perpendicular, if a₁a₂+ b₁b₂+ c₁c₂= 0
$\therefore$ -3 (3k) + 2k × 1 + 2 (-5) = 0
⇒ -9k + 2k - 10 = 0
⇒ 7k = - 10
$\Rightarrow\text{k}=\frac{-10}{7}$
Therefore, for $\text{k}=-\frac{10}{7},$ the given lines are perpendicular to each other.

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Question 223 Marks

Find the vector equation of the line passing through the point A(1, 2, -1) and parallel to the line 5x - 25 = 14 - 7y = 35z.

Answer

The equation of the line 5x - 25 = 14 - 7y = 35z can be re-written as
$\frac{\text{x}-5}{\frac{1}{5}}=\frac{\text{y}-2}{\frac{-1}{7}}=\frac{\text{z}}{\frac{1}{35}}$
$\Rightarrow\frac{\text{x}-5}{7}=\frac{\text{y}-2}{-5}=\frac{\text{z}}{1}$
Since the required line is parallel to the given line, so the direction ration of the required line are proportional to 7, -5, 1.
The vector equation of the required line passing through the point (1, 2, -1) and having direction ratios proportional to 7, -5, 1 is
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).$

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Question 233 Marks

Show that the line joining the origin to the points (2, 1, 1) is perpendicular to the line detarmined by the points (3, 5, -1) and (4, 3, -1).

Answer

The direction ratios of the line joining the origin to the point (2, 1, 1) are 2, 1, 1.
Let $\vec{\text{b}}_1=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
The direction ratios ot the line joining the points (3, 5, -1) and (4, 3, -1) are 1, -2, 0.
Let $\vec{\text{b}}_2=\hat{\text{i}}-2\hat{\text{j}}+0\hat{\text{k}}$
Now,
$\vec{\text{b}}_1.\vec{\text{b}}_2=\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+0\hat{\text{k}}\big)$
$=2-2+0$.
$=0$
$\therefore\vec{\text{b}}_1\perp\vec{\text{b}}_2$
Hence, the two lines joining the given points are perpendicular to each other.

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Question 243 Marks

Write the value of $\lambda$ for which the lines $\frac{\text{x}-3}{-3}=\frac{\text{y}+2}{2\lambda}=\frac{\text{z}+4}{2}$ and $\frac{\text{x}+1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}+6}{-5}$ are perpendicular to each other.

Answer

We have
$\frac{\text{x}-3}{-3}=\frac{\text{y}+2}{2\lambda}=\frac{\text{z}+4}{2}$
$\frac{\text{x}+1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}+6}{-5}$
The given lines are parallel to vector $\vec{\text{b}}_1=-3\hat{\text{i}}+2\lambda\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}_2=3\lambda\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}.$
For $\vec{\text{b}}_1\perp\vec{\text{b}}_2,$ we must have
$\vec{\text{b}}_1.\vec{\text{b}}_2=0$
$\Rightarrow\big(-3\hat{\text{i}}+2\lambda\hat{\text{j}}+2\hat{\text{k}}\big).\big(3\lambda\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}\big)=0$
$\Rightarrow-7\lambda-10=0$
$\Rightarrow\lambda=-\frac{10}7{}$

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Question 253 Marks

Find the cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by $\frac{\text{x}+3}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+8}{6}.$

Answer

We know that the cartesian equation of a line passing with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_2}{\text{b}}=\frac{\text{z}-\text{z}_3}{\text{c}}.$
Here,
$\vec{\text{a}}=-2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}$
The cartesian equation of the required line is
$\frac{\text{x}-(-2)}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}-(-5)}{6}$
$=\frac{\text{x}+2}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+5}{6}$

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Question 263 Marks

Find the vector equation of the line passing through the point (2, -1, -1) which is parallel to the line 6x - 2 = 3y +1 =2z - 2.

Answer

The equation of the line 6x - 2 = 3y + 1 = 2z - 2 can be re-written as
$\frac{\text{x}-\frac{1}{3}}{\frac{1}{6}}=\frac{\text{y}+\frac{1}{3}}{\frac{1}{3}}=\frac{\text{z}-1}{\frac{1}{2}}$
$=\frac{\text{x}-\frac{1}{3}}{1}=\frac{\text{y}+\frac{1}{3}}{2}=\frac{\text{z}-1}{3}$
Since the reqired line is parallel to the given line, the direction ratios of the recuired line are proportional to 1, 2, 3.
The vector equation of the required line passing through the point (2, -1, -1) and having direction ratios proportional to 1, 2, 3 is $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big).$

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Question 273 Marks

Find the perpendicular distence of the point (3, -1, 11) from the line $\frac{\text{x}}{2}=\frac{\text{y}-2}{-3}=\frac{\text{z}-3}{4}.$

Answer

Let the point (3, -1, 11) be P and the point through which the line passes be Q (0, 2, 3).
The line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
Now,
$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&4\\-3&3&-8\end{vmatrix}$
$=12\hat{\text{i}}+4\hat{\text{j}}+15\hat{\text{k}}$
$\Rightarrow|\vec{\text{b}}\times\overrightarrow{\text{PQ}}|=\sqrt{12^2+4^2+15^2}$
$=\sqrt{144+16+225}$
$=\sqrt{385}$
$\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{385}}{\sqrt{29}}$
$=\frac{\sqrt{385}}{\sqrt{29}}$

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Question 283 Marks

Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).

Answer

The direction ratios of the line joining the points (3, 1, 4) and (7, 2, 12) are proportional to 4, 1, 8.
Let $\vec{\text{m}}_1$ and $\vec{\text{m}}_2$ be vectors parallel to the lines having direction ratios proportional to 2, 2, 1 and 4, 1, 8.
Now,
$\vec{\text{b}}_1=2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}_2=4\hat{\text{i}}+\hat{\text{j}}+8\hat{\text{k}}$
If $\theta$ is the angle between the given lines, then
$\cos\theta=\frac{\vec{\text{m}}_1.\vec{\text{m}}_2}{|\vec{\text{m}}_1||\vec{\text{m}}_2|}$
$=\frac{\big(2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big).\big(4\hat{\text{i}}+\hat{\text{j}}+8\hat{\text{k}}\big)}{\sqrt{2^2+2^2+1^2}\sqrt{4^2+1^2+8^2}}$
$=\frac{8+2+8}{3\times9}$
$=\frac{2}{3}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{2}{3}\big)$

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Question 293 Marks

Write the angle between the lines 2x = 3y = -z and 6x = -y = -4z.

Answer

We have
2x = 3y = -z
6x = -y = -4z
The given lines can be re-written as
$\frac{\text{x}}{\frac{1}{2}}=\frac{\text{y}}{\frac{1}{3}}=\frac{\text{z}}{-1}$ and $\frac{\text{x}}{\frac{1}{6}}=\frac{\text{y}}{-1}=\frac{\text{z}}{-\frac{1}{4}}$
$\Rightarrow\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$ and $\frac{\text{x}}{2}=\frac{\text{y}}{-12}=\frac{\text{z}}{-3}$
These lines are parallel to vectors $\vec{\text{b}}_1=3\hat{\text{i}}+2\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}}_2=2\hat{\text{i}}-12\hat{\text{j}}-3\hat{\text{k}}$
Let $\theta$ be the angle between these lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(3\hat{\text{i}}+2\hat{\text{j}}-6\hat{\text{k}}\big).\big(2\hat{\text{i}}-12\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{3^2+2^2+(-6)^2}\sqrt{2^2+(-12)^2+(-3)^2}}$
$=\frac{6-24+18}{\sqrt{9+4+36}\sqrt{4+144+9}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$

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Question 303 Marks

Find the vector equation of a line passing through (2, -1, 1) and parallel to the line whose equations are $\frac{\text{x}-3}{2}=\frac{\text{y}+1}{7}=\frac{\text{z}-2}{-3}.$

Answer

We know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
Here,
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}$
Vector equation of the required line is
$\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.

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