2 Marks

Question 12 Marks

Write the vector equation of a line passing through a point having position vector $\vec{\alpha}$ and parallel to vector $\vec{\beta}.$

Answer

The vector equation of the line passing the point having position vector $\vec{\alpha}$ and parallel to vector $\vec{\beta}$is $\vec{\text{r}}=\vec{\alpha}+\lambda\vec{\beta}.$

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Question 22 Marks

Write the formula for the shortest distance between the lines

$\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}.$

Answer

The shortest distance d between the parallel lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}$ is given by
$\text{d}=\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|}{\big|\vec{\text{b}}\big|}$

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Question 32 Marks

Write the vector equation of a line given by $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$

Answer

We have
$\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}$
The given line passes through the point (5, -4, 6) and has direction ratios proportional to 3, 7, 2.
Vector equation of the given line passing through the point having position vector $\vec{\text{a}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}$ and parallel to a vector $\vec{\text{b}}=3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}$ is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda\big(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}\big)$

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Question 42 Marks

Write the direction consines of the line whose cartesian equations are 2x = 3y = -z.

Answer

We have

2x = 3y = -z

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

$\frac{\text{x}}{\frac{1}{2}}=\frac{\text{y}}{\frac{1}{3}}=\frac{\text{z}}{-1}$

$\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$

The diraction ratios of the line parallel to AB are proportional to 3, 2, -6.

Hence, the direction cosines of the line parallel to AB are proportional to

$\frac{3}{\sqrt{3^2+2^2+(-6)^2}},\frac{2}{\sqrt{3^2+2^2+(-6)^2}},\frac{-6}{\sqrt{3^2+2^2+(-6)^2}}$

$=\frac{3}{7},\frac{2}{7},-\frac{6}{7}$

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Question 52 Marks

The equations of a line are given by $\frac{4-\text{x}}{3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}.$ Write the direction cosines of a line parallel to this line.

Answer

We have

$\frac{4-\text{x}}{3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}$

The equation of the given line can be re-written as.

$\frac{\text{x}-4}{-3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}$

The direction ratios of the line parallel to the given line are proportional to -3, 3, 6.

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

$\frac{-3}{\sqrt{(-3)^2+3^2+6^2}},\frac{-3}{\sqrt{(3)^2+3^2+6^2}},\frac{6}{\sqrt{(-3)^2+3^2+6^2}}$

$=\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}$

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Question 62 Marks

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

Answer

The equation of the given line is $\frac{\text{x}+3}{3}=\frac{4-\text{y}}{5}=\frac{\text{z}+8}{6}$

It can be re-written as

$\frac{\text{x}+3}{3}=\frac{\text{y}-4}{-5}=\frac{\text{z}+8}{6}$

Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 3, -5, 6.

Hence, the cartesian equations of the line passing through the point (-2, 4, -5) and parallel to a vector having direction ratios proportional to 3, -5, 6 is $\frac{\text{x}+2}{3}=\frac{\text{y}-4}{-5}=\frac{\text{z}+5}{6}.$

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Question 72 Marks

Find the equation of a line parallel to x-axis and passing through the origin.

Answer

Vector equation of a line is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
The direction cosines of the x-axis are (1, 0, 0). Equation of a line parallel to the x-axis and passing through the origin is
$\vec{\text{r}}=\big(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)+\lambda\big(1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)$
$\vec{\text{r}}=\lambda\hat{\text{i}}$

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Question 82 Marks

Find the equation of the line passing through the points (2, -1, 3) and parallel to the line $\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}\big).$

Answer

The given line is parallel to the vector $2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$ and the required line is parallel to the given line.
So, the required line is parallel to the vector $2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
hence, the equation of the required line passing through the points (2, -1, 3) and parallel to the vector
$2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$ is $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}\big)$

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Question 92 Marks

Find the angle between the lines $\vec{\text{r}}=\big(2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$ and $\vec{\text{r}}=7\hat{\text{i}}-6\hat{\text{k}}+\mu\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$

Answer

Let $\theta$ be the angle between the given lines. The given lines are parallel to the vectors $\vec{\text{b}}_1=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}},$ respectively.
So, the angle $\theta$ between the given lines is given by
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)}{\sqrt{3^2+2^2+6^2}\sqrt{1^2+2^+2^2}}$
$=\frac{3\times1+2\times2+6\times2}{\sqrt{49}\sqrt{9}}$
$=\frac{19}{21}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{19}{21}\big)$
Thus, the angle between the given lines is $\cos^{-1}\big(\frac{19}{21}\big).$

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Question 102 Marks

If the equations of a line AB are $\frac{3-\text{x}}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-5}{4},$ write the direction ratios of a line parallel to AB.

Answer

We have

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

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

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

Thus, the direction ratios of the line parllel to Ab are proportional to -1, -2, 4.

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Question 112 Marks

Write the condition for the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ to be intersecting.

Answer

The shortest distance d between the parallel lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
For the lines to be intersecting, d = 0.
$\Rightarrow\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}=0$
$\Rightarrow\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$

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Question 122 Marks

Write the cartesian and vector equations of x-axis.

Answer

Since x-axis passes through the point (0, 0, 0) having position vector $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ having direction ratios proportional to 1, 0, 0, the cartesian equation of x-axis is
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{0}$
$=\frac{\text{x}}{1}=\frac{\text{y}}{0}=\frac{\text{z}}{0}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)$
$=\lambda\hat{\text{i}}$

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Question 132 Marks

Write the coordinate axis to which the line $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-1}{0}$ is perpendicular.

Answer

We have
$\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-1}{0}$
The given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+0\hat{\text{k}}$
Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be perpendicular to the given line.
Now,
$3\text{x}+4\text{y}+0\text{z}=0$
It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).
Hence, the given line is perpendicular to z-axis.

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Question 142 Marks

Write the cartesian and vector equations of z-axis.

Answer

Since z-axis passes through the point (0, 0, 0) having position vector $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=0\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}$ having direction ratios proportional to 0, 0, 1, the cartesian equation of z-axis is
$\frac{\text{x}-0}{0}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{1}$
$=\frac{\text{x}}{0}=\frac{\text{y}}{0}=\frac{\text{z}}{1}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(0\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}\big)$
$=\lambda\hat{\text{k}}$

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Question 152 Marks

Write the cartesian and vector equations of y-axis.

Answer

Since y-axis passes through the point (0, 0, 0) having position vector $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=0\hat{\text{i}}+1\hat{\text{j}}+0\hat{\text{k}}$ having direction ratios proportional to 0, 1, 0, the cartesian equation of y-axis is
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{0}$
$=\frac{\text{x}}{0}=\frac{\text{y}}{1}=\frac{\text{z}}{0}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(0\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}\big)$
$=\lambda\hat{\text{j}}$

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