RD Sharma Class 12 Ex 28.2 Solutions Chapter 28 The Straight Line in Space

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter28
Exercise28.2
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 28.2 Solutions Chapter 28 The Straight Line in Space

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Show that the three lines with direction cosines 12/13, -3/13, – 4/13; 4/13, 12/13, 3/13; 3/13, – 4/13, 12/13 are mutually perpendicular.

Solution:

The direction cosines of the three lines are

l1 = 12/13, m1 = -3/13, n1 = -4/13

l2 = 4/13, m2 = 12/13, n2 = 3/13

l3 = 3/13, m= -4/13, n= 12/13

So, l1 l2 + m1 m2 + n1 n2 =\frac{48 - 36 - 12}{169}  = 0

Also,

l2 l3 + m2 m3 + n2 n3 =\frac{12 - 48 + 36}{169}  = 0

l1 l3 + m1 m3 + n1 n3 =\frac{36 + 12 - 48}{169}  = 0

Therefore, the given lines are perpendicular to each other.

Hence proved.

Question 2. 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).

Solution:

We have,

\vec{a}  is passing through the points (1, -1, 2) and (3, 4, -2).

Also,\vec{a}  is passing through the points (0, 3, 2) and (3, 5, 6).

Then,

\vec{a} = 2 \hat{i} + 5 \hat{j} - 4\hat{k}
\vec{b} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}

Now,

\vec{a} . \vec{b} = \left( 2 \hat{i} + 5 \hat{j} - 4 \hat{k} \right) . \left( 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)

= 6 + 10 – 16

= 0

Therefore, the given lines are perpendicular to each other.

Hence proved.

Question 3. 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).

Solution:

Equations of lines passing through the points (x1, y1, z1) and (x2, y2, z2) are given by

\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}

So, the equation of a line passing through (4, 7, 8) and (2, 3, 4) is

\frac{x - 4}{2 - 4} = \frac{y - 7}{3 - 7} = \frac{z - 8}{4 - 8}
\frac{x - 4}{- 2} = \frac{y - 7}{- 4} = \frac{z - 8}{- 4}

Also, the equation of the line passing through the points ( -1, -2,1) and (1, 2, 5) is

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

We know that two lines are parallel if,

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

And the Cartesian equations of the two lines are given by,

\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}
\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}

So, we get,

\frac{- 2}{2} = \frac{- 4}{4} = \frac{- 4}{4} = -1

Therefore, the given lines are parallel to each other.

Hence proved.

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

Solution:

We know that the cartesian equation of a line passing through a point with position vector\vec{a}  and parallel to the vector\vec{b}  is given by,

\frac{x - x_1}{a} = \frac{y - y_2}{b} = \frac{z - z_3}{c}

Here,

\vec{a} = -2 \hat{i} + 4j - 5 \hat{k}
\vec{b} = 3 \hat{i} + 5 \hat{j} - 6 \hat{k}

The cartesian equation of the required line is,

\frac{x - \left( - 2 \right)}{3} = \frac{y - 4}{5} = \frac{z - \left( - 5 \right)}{6}

=>\frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}

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

Solution:

We have

\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1}

And also,

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

These equations can be re-written as,

\frac{x - 5}{7} = \frac{y - \left( - 2 \right)}{- 5} = \frac{z - 0}{1}  . . . . (1)

\frac{x - 0}{1} = \frac{y - 0}{2} = \frac{z - 0}{3}  . . . . (2)

Therefore, the vector parallel to line (1) is given by,

\vec{m_1} = 7 \hat{i} - 5 \hat{j} + \hat{k}

And the vector parallel to line (2) is given by,

\vec{m_2} = \hat{i} + 2 \hat{j} + 3 \hat{k}

Now,

\vec{m_1} . \vec{m_2} = \left( 7 \hat{i} - 5 \hat{j} + \hat{k} \right) . \left( \hat{i}+ 2 \hat{j} + 3 \hat{k} \right)

= 7 – 10 + 3

= 0

Therefore, the given two lines are perpendicular to each other.

Hence proved.

Question 6. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1).

Solution:

The direction ratios of the line joining the origin to the point (2, 1, 1) are 2, 1, 1.

Let\vec{b_1} = 2 \hat{i} + \hat{j} + \hat{k}

The direction ratios of the line joining the points (3, 5, -1) and (4, 3, -1) are 1, -2, 0.

Let\vec{b_2} = \hat{i} - 2 \hat{j} + 0 \hat{k}

Now,

\vec{b_1} . \vec{b_2} = \left( 2 \hat{i} + \hat{j} + \hat{k } \right) . \left( \hat{i} - 2 \hat{j} + 0 \hat{k} \right)

= 2 – 2 + 0

= 0

So, we get\overrightarrow{b_1} \perp \overrightarrow{b_2}  .

Therefore, the two lines joining the given points are perpendicular to each other.

Hence proved.

Question 7. Find the equation of a line parallel to x-axis and passing through the origin.

Solution:

The direction ratios of the line parallel to x-axis are proportional to 1, 0, 0.

Equation of the line passing through the origin (0, 0, 0) and parallel to x-axis is

\frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{0}

=>\frac{x}{1} = \frac{y}{0} = \frac{z}{0}

Question 8. Find the angle between the following pair of line:

(i)\vec{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right)  and\vec{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)

Solution:

We have,

\vec{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right)

And also,

\vec{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)

Let\vec{b_1}  and\vec{b_2}  be vectors parallel to the given lines .

Now,

\vec{b_1} = \hat{i} + 2 \hat{j} - 2 \hat{k}
\vec{b_2} = 2 \hat{i} + 4 \hat{j} - 4 \hat{k}

If θ is the angle between the given lines, then

\cos \theta = \frac{\vec{b_1} . \vec{b_2}}{\left| \vec{b_1} \right| \left| \vec{b_2} \right|}

=\frac{\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) . \left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)}{\sqrt{1^2 + 2^2 + \left( - 2 \right)^2} \sqrt{2^2 + 4^2 + \left( - 4 \right)^2}}

=\frac{2 + 8 + 8}{3(6)}

= 1

As cos θ = 1

=> θ = 0°

Therefore, the angle between two lines is 0°.

(ii)\vec{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right)  and\vec{r} = \left( 5 \hat{j} - 2 \hat{k} \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)

Solution:

We have,

\vec{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right)

And also,

\vec{r} = \left( 5 \hat{j} - 2 \hat{k} \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)

Let\vec{b_1}  and\vec{b_2}  be vectors parallel to the given lines .

Now,

\vec{b_1} = \hat{i} + 2 \hat{j} + 2 \hat{k}
\overrightarrow{b_2} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}

If θ is the angle between the given line, then

\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}

=\frac{\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) . \left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{3^2 + 2^2 + 6^2}}

=\frac{3 + 4 + 12}{3 \times 7}

= 19/21

As cos θ = 19/21

=> θ = cos-1 (19/21)

Therefore, the angle between two lines is cos-1 (19/21).

(iii)\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right)  and\overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}

Solution:

We have,

\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right)

And also,

\overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}

Let\overrightarrow {b_1}  and\overrightarrow {b_2}  be vector parallel to the given line.

Now,

\overrightarrow{b_1} = \hat{i} + \hat{j} + 2 \hat{k}
\overrightarrow{b_2} = \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k}

If θ is the angle between the given line, then

\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}

=\frac{\left( \hat{i} + \hat{j} + 2 \hat{k} \right) . \left( \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right)}{\sqrt{1^2 + 1^2 + 2^2} \sqrt{\left( \sqrt{3} - 1 \right)^2 + \left( \sqrt{3} + 1 \right)^2 + 4^2}}

=\frac{\left( \sqrt{3} - 1 \right) - \left( \sqrt{3} + 1 \right) + 8}{\sqrt{6} \sqrt{24}}

= 6/12

= 1/2

As cos θ = 1/2

=> θ = π/3

Therefore, the angle between two lines is π/3.

Question 9. Find the angle between the following pair of line:

(i)\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4}  and \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}

Solution:

We have,

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

And also,

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

Let,\overrightarrow{b_1}  and\overrightarrow{b_2}  be vectors parallel to the given line.

\overrightarrow{b_1} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}
\overrightarrow{b_2} = \hat{i} + \hat{j}+ 2 \hat{k}

If θ is the angle between the given line, then

\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}

=\frac{\left( 3 \hat{i} + 5 \hat{j} + 4 \hat{k} \right) . \left( \hat{i} + \hat{j} + 2 \hat{k} \right)}{\sqrt{3^2 + 5^2 + 4^2} \sqrt{1^2 + 1^2 + 2^2}}

=\frac{3 + 5 + 8}{10\sqrt{3}}

= 8/5√3

As cos θ = 8/5√3

=> θ = cos-1 (8/5√3)

Therefore, the angle between two lines is cos-1 (8/5√3).

(ii)\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3}  and\frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

Solution:

We have,

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

And also,

\frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

Let\overrightarrow{b_1}  and\overrightarrow{b_2}  be vectors parallel to the given lines.

Now,

\overrightarrow{b_1} = 2 \hat{i} + 3 \hat{j} - 3 \hat{k}
\overrightarrow{b_2} = - \hat{i} + 8 \hat{j} + 4 \hat{k}

If θ is the angle between the given lines, then

\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}

=\frac{\left( 2 \hat{i} + 3 \hat{j} - 3 \hat{k} \right) . \left( - \hat{i} + 8 \hat{j} + 4 \hat{k} \right)}{\sqrt{2^2 + 3^2 + \left( - 3 \right)^2} \sqrt{\left( - 1 \right)^2 + 8^2 + 4^2}}

=\frac{- 2 + 24 - 12}{9\sqrt{22}}

=\frac{10}{9\sqrt{22}}

As cos θ =\frac{10}{9\sqrt{22}}

=> θ =\cos^{- 1} \left( \frac{10}{9\sqrt{22}} \right)

Therefore, the angle between two lines is\cos^{- 1} \left( \frac{10}{9\sqrt{22}} \right)  .

(iii)\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3}  and\frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}

Solution:

We have,

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

And also,

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

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

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

Let\overrightarrow{b_1}  and\overrightarrow{b_2}  be vectors parallel to the given lines.

Now,

\overrightarrow{b_1} = 2 \hat{i} + \hat{j} - 3 \hat{k}
\overrightarrow{b_2} = 3 \hat{i} + 2 \hat{j} - \hat{k}

If θ is the angle between the given lines, then

\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}

=\frac{\left( 2 \hat{i} + \hat{j} - 3 \hat{k} \right) . \left( 3 \hat{i} + 2 \hat{j} - \hat{k} \right)}{\sqrt{2^2 + 1^2 + \left( - 3 \right)^2} \sqrt{3^2 + 2^2 + \left( - 1 \right)^2}}

=\frac{6 + 2 + 3}{\sqrt{14} \sqrt{14}}

= 11/14

As cos θ = 11/14

=> θ = cos-1 (11/14)

Therefore, the angle between two lines is cos-1 (11/14).

(iv)\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5  and\frac{x + 1}{1} = \frac{2y - 3}{3} = \frac{z - 5}{2}

Solution:

We have,

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

And also,

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

The equations of the given lines can be re-written as

\frac{x - 2}{3} = \frac{y + 3}{- 2} = \frac{z - 5}{0}
\frac{x + 1}{1} = \frac{y - \frac{3}{2}}{\frac{3}{2}} = \frac{z - 5}{2}

Let\overrightarrow{ b_1}  and\overrightarrow{ b_2}  be vectors parallel to the given lines.

Now,

\overrightarrow{b_1} = 3 \hat{i} - 2 \hat{j} + 0 \hat{k}
\overrightarrow{b_2} = \hat{i} + \frac{3}{2} \hat{j} + 2 \hat{k}

If θ is the angle between the given lines, then

\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}

=\frac{\left( 3 \hat{i} - 2 \hat{j} + 0 \hat{k} \right) . \left( \hat{i} + \frac{3}{2} \hat{j} + 2 \hat{k} \right)}{\sqrt{3^2 + \left( - 2 \right)^2 + 0^2} \sqrt{1^2 + \left( \frac{3}{2} \right)^2 + 2^2}}

=\frac{3 - 3 + 0}{\sqrt{13} \sqrt{\frac{29}{4}}}

= 0

As cos θ = 0

=> θ = π/2

Therefore, the angle between two lines is π/2.

(v)\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1}  and\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

Solution:

We have,

\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1}

And also,

\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

The equations of the given lines can be re-written as,

\frac{x - 5}{1} = \frac{y + 3}{- 1} = \frac{z - 3}{1}
\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

Let\overrightarrow { b_1}  and\overrightarrow { b_2}  be vectors parallel to the given lines.

Now,

\overrightarrow{b_1} = \hat{i} - \hat{j} + \hat{k}
\overrightarrow{b_2} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}

If θ is the angle between the given lines, then

\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}

=\frac{\left( \hat{i} - \hat{j} + \hat{k} \right) . \left( 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \right)}{\sqrt{1^2 + \left( - 1 \right)^2 + 1^2} \sqrt{3^2 + 4^2 + 5^2}}

=\frac{3 - 4 + 5}{\sqrt{3} \sqrt{50}}

= 4/5√6

As cos θ = 4/5√6

=> θ = cos-1 (4/5√6)

Therefore, the angle between two lines is cos-1 (4/5√6).

(vi)\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3}  and\frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}

Solution:

We have,

\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3}

And also,

\frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}

The equations of the given lines can be re-written as

\frac{x - 2}{2} = \frac{y - 1}{7} = \frac{z + 3}{- 3}
\frac{x + 2}{- 1} = \frac{y - 4}{2} = \frac{z - 5}{4}

Let\overrightarrow{b_1}  and\overrightarrow{b_2}  be vectors parallel to the given lines.

Now,

\overrightarrow{b_1} = 2 \hat{i} + 7 \hat{j} - 3 \hat{k}
\overrightarrow{b_2} = - 1 \hat{i} + 2 \hat{j} + 4 \hat{k}

If θ is the angle between the given lines, then

\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}

=\frac{\left( 2 \hat{i} + 7 \hat{j} - 3 \hat{k} \right) . \left( - 1 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)}{\sqrt{2^2 + 7^2 + \left( - 3 \right)^2} \sqrt{\left( - 1 \right)^2 + 2^2 + 4^2}}

=\frac{- 2 + 14 - 12}{\sqrt{62} \sqrt{21}}

= 0

As cos θ = 0

=> θ = π/2

Therefore, the angle between two lines is π/2.

Question 10. Find the angle between the pairs of lines with direction ratios proportional to:

(i) 5, −12, 13 and −3, 4, 5

Solution:

We have pairs of lines with direction ratios proportional to 5, −12, 13 and −3, 4, 5.

Let\overrightarrow{m_1}  and\overrightarrow{m_2}  be vectors parallel to the two given lines.

Then, the angle between the two given lines is same as the angle between\overrightarrow{m_1}  and\overrightarrow{m_2}  .

Now,

The vector parallel to the line having direction ratios proportional to 5, – 12, 13 is,

\overrightarrow{m_1} = 5 \hat{i} - 12 \hat{j} + 13 \hat{k}

And the vector parallel to the line having direction ratios proportional to -3, 4, 5 is,

\overrightarrow{m_2} = - 3 \hat{i} + 4 \hat{j} + 5 \hat{k}

Let θ be the angle between the lines.

Now,

\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}

=\frac{\left( 5 \hat{i} - 12 \hat{j} + 13 \hat{k} \right) . \left( - 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \right)}{\sqrt{5^2 + \left( - 12 \right)^2 + {13}^2} \sqrt{\left( - 3 \right)^2 + 4^2 + 5^2}}

=\frac{- 15 - 48 + 65}{13\sqrt{2} \times 5\sqrt{2}}

= 1/65

As cos θ = 1/65

=> θ = cos-1 (1/65)

Therefore, the angle between two lines is cos-1 (1/65).

(ii) 2, 2, 1 and 4, 1, 8

Solution:

We have pairs of lines with direction ratios proportional to 2, 2, 1 and 4, 1, 8.

Let\overrightarrow{m_1}  and\overrightarrow{m_2}  be vectors parallel to the given two lines.

Then, the angle between the lines is same as the angle between\overrightarrow{m_1}  and\overrightarrow{m_2}  .

Now,

The vector parallel to the line having direction ratios proportional to 2, 2, 1 is,

\overrightarrow{m_1} = 2 \hat{i} - 2 \hat{j} + \hat{k}

And the vector parallel to the line having direction ratios proportional to 4, 1, 8 is,

\overrightarrow{m_2} = 4 \hat{i} + \hat{j} + 8 \hat{k}

Let θ be the angle between the lines.

Now,

\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}

=\frac{\left( 2 \hat{i} + 2 \hat{j} + \hat{k} \right) . \left( 4 \hat{i}+ \hat{j} + 8 \hat{k} \right)}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{4^2 + 1^2 + 8^2}}

=\frac{8 + 2 + 8}{3 \times 9}

= 2/3

As cos θ = 2/3

=> θ = cos-1 (2/3)

Therefore, the angle between two lines is cos-1 (2/3).

(iii) 1, 2, −2 and −2, 2, 1

Solution:

We have pairs of lines with direction ratios proportional to 1, 2, −2 and −2, 2, 1.

Let\overrightarrow{m_1}  and\overrightarrow{m_2}  be vectors parallel to the two given lines.

Then, the angle between the two given lines is same as the angle between\overrightarrow{m_1}  and\overrightarrow{m_2}  .

Now,

The vector parallel to the line having direction ratios proportional to 1, 2, – 2 is,

\overrightarrow{m_1} = \hat{i} + 2 \hat{j} - 2 \hat{k}

And the vector parallel to the line having direction ratios proportional to -2, 2, 1 is,

\overrightarrow{m_2} = - 2 \hat{i} + 2 \hat{j} + \hat{k}

Let θ be the angle between the lines.

Now,

\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}

=\frac{\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) . \left( - 2 \hat{i} + 2 \hat{j} + \hat{k} \right)}{\sqrt{1^2 + 2^2 + \left( - 2 \right)^2} \sqrt{\left( - 2 \right)^2 + 2^2 + 1^2}}

=\frac{- 2 + 4 - 2}{3 \times 3}

= 0

As cos θ = 0

=> θ = π/2

Therefore, the angle between two lines is π/2.

(iv) a, b, c and b − c, c − a, a − b

Solution:

We have pairs of lines with direction ratios proportional to a, b, c and b − c, c − a, a − b.

Let\overrightarrow{m_1}  and\overrightarrow{m_2}  be vectors parallel to the given two lines.

Then, the angle between the two lines is same as the angle between\overrightarrow{m_1}  and\overrightarrow{m_2}  .

Now,

The vector parallel to the line having direction ratios proportional to a, b, c is,

\overrightarrow{m_1} = a \hat{i} + b \hat{j} + c \hat{k}

And the vector parallel to the line having direction ratios proportional to b – c, c – a, a – b is,

\overrightarrow{m_2} = \left( b - c \right) \hat{ i }+ \left( c - a \right) \hat{j} + \left( a - b \right) \hat{k}

Let θ be the angle between the lines.

Now,

\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}

=\frac{\left( a \hat{i} + b \hat{j} + c \hat{k} \right) . \left\{ \left( b - c \right) \hat{i} + \left( c - a \right) \hat{j} + \left( a - b \right) \hat{k} \right\}}{\sqrt{a^2 + b^2 + c^2} \sqrt{\left( b - c \right)^2 + \left( c - a \right)^2 + \left( a - b \right)^2}}

=\frac{ab - ac + bc - ba + ca - cb}{\sqrt{a^2 + b^2 + c^2} \sqrt{\left( b - c \right)^2 + \left( c - a \right)^2 + \left( a - b \right)^2}}

= 0

As cos θ = 0

=> θ = π/2

Therefore, the angle between two lines is π/2.

Question 11. 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).

Solution:

The direction ratios of the line joining the points (3, 1, 4) and (7, 2, 12) are proportional to 4, 1, 8.

Let\overrightarrow{m_1}  and\overrightarrow{m_2}  be vectors parallel to the lines having direction ratios proportional to 2, 2, 1 and 4, 1, 8.

Now,

\overrightarrow{b_1} = 2 \hat{i} + 2 \hat{j} + \hat{k}
\overrightarrow{b_2} = 4 \hat{i} + \hat{j} + 8 \hat{k}

If θ is the angle between the given lines, then

\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}

=\frac{\left( 2 \hat{i} + 2 \hat{j} + \hat{k} \right) . \left( 4 \hat{i} + \hat{j} + 8 \hat{k} \right)}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{4^2 + 1^2 + 8^2}}

=\frac{8 + 2 + 8}{3 \times 9}

= 2/3

As cos θ = 2/3

=> θ = cos-1 (2/3)

Therefore, the angle between two lines is cos-1 (2/3).

Question 12. Find the equation of the line passing through the point (1, 2, −4) and parallel to the line\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{z + 1}{3}  .

Solution:

The direction ratios of the line parallel to line\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{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{x - 1}{4} = \frac{y - 2}{2} = \frac{z - \left( - 4 \right)}{3}

=>\frac{x - 1}{4} = \frac{y - 2}{2} = \frac{z + 4}{3}

Question 13. Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line \frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .

Solution:

The equation of line \frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3}  can be re-written as,

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

The direction ratios of the line parallel to line \frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3}  are proportional to 2, 2/3, -3.

Equation of the required line passing through the point ( -1, 2, 1) having direction ratios proportional to (2, 2/3, -3) is,

\frac{x - \left( - 1 \right)}{2} = \frac{y - 2}{\frac{2}{3}} = \frac{z - 1}{- 3}

=> \frac{x + 1}{2} = \frac{y - 2}{\frac{2}{3}} = \frac{z - 1}{- 3}

Question 14. Find the equation of the line passing through the point (2, −1, 3) and parallel to the line \overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .

Solution:

The given line \overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right)  is parallel to the vector 2 \hat{i} + 3 \hat{j} - 5 \hat{k}

And the required line is also parallel to the given line. 

So, the required line is parallel to the vector 2 \hat{i} + 3 \hat{j} - 5 \hat{k}

Hence, the equation of the required line passing through the point (2,-1, 3) and parallel to the vector  2 \hat{i} + 3 \hat{j} - 5 \hat{k}  is, 

=> \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right)

Question 15. Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}  and \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5} .

Solution:

Let,

\overrightarrow{b_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}
\overrightarrow{b_2} = - 3 \hat{i} + 2 \hat{j} + 5 \hat{k}

Since the required line is perpendicular to the lines parallel to the vectors  \overrightarrow{b_1}  and \overrightarrow{b_2} , it is also parallel to the vector \overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}   

Now,  

\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}

\begin{vmatrix}\hat{i} & \hat{j}  & \hat{k} \\ 1 & 2 & 3 \\ - 3 & 2 & 5\end{vmatrix}

4 \hat{i} - 14 \hat{j} + 8 \hat{k}

2\left( 2 \hat{i} - 7 \hat{j} + 4 \hat{k}  \right)

Thus, the direction 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{x - 2}{2} = \frac{y - 1}{- 7} = \frac{z - 3}{4}

Question 16. Find the equation of the line passing through the point \hat{i}  + \hat{j}  - 3 \hat{k}  and perpendicular to the lines \overrightarrow{r} = \hat{i}  + \lambda\left( 2 \hat{i} + \hat{j}  - 3 \hat{k}  \right)  and \overrightarrow{r} = \left( 2 \hat{i}  + \hat{j}  - \hat{ k}  \right) + \mu\left( \hat{i}  + \hat{j}  + \hat{k}  \right) .

Solution:

The required line is perpendicular to the lines parallel to the vectors \overrightarrow{b_1} = 2 \hat{i} + \hat{j} - 3 \hat{k}  and \overrightarrow{b_2} = \hat{ i} + \hat{j}+ \hat{k} .

So, the required line is parallel to the vector,

\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}

\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 3 \\ 1 & 1 & 1\end{vmatrix}

4 \hat{i} - 5 \hat{j} + \hat{k}

Equation of the required line passing through the point \left( \hat{i} + \hat{j} - 3 \hat{k}  \right)  and parallel to \left( 4 \hat{i}  - 5 \hat{j} + \hat{k}  \right)  is,

=> \overrightarrow{r} = \left( \hat{i}  + \hat{j} - 3 \hat{k} \right) + \lambda\left( 4 \hat{i} - 5 \hat{j}  + \hat{k} \right)

Question 17. Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0), and (1, 2, −1), (2, 1, 1).

Solution:

The direction ratios of the line joining the points (4, 3, 2), (1, -1, 0) and (1, 2, -1), (2, 1, 1) are -3, -4, -2 and 1, -1, 2 respectively.  

Let, 

\overrightarrow{b_1} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k}
\overrightarrow{b_2} = \hat{i} - \hat{j} + 2 \hat{k}

Since the required line is perpendicular to the lines parallel to the vectors \overrightarrow{b_1} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k}  and \overrightarrow{b_2} = \hat{i} - \hat{j} + 2 \hat{k} , it is parallel to the vector \overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}

Now,

\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}

\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ - 3 & - 4 & - 2 \\ 1 & - 1 & 2\end{vmatrix}

-10 \hat{i} + 4 \hat{j} + 7 \hat{k}

So, the direction ratios of the required line are proportional to -10, 4, 7.

The equation of the required line passing through the point (1,-1, 1) and having direction ratios proportional to -10, 4, 7 is  

=> \frac{x - 1}{- 10} = \frac{y + 1}{4} = \frac{z - 1}{7}

Question 18. Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7}  and \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5} .

Solution:

We have,

\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7}
\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}

Let,

\overrightarrow{b_1} = 8 \hat{i} - 16 \hat{j} + 7 \hat{k}
\overrightarrow{b_2} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k}

Since the required line is perpendicular to the lines parallel to the vectors \overrightarrow{b_1}  and \overrightarrow{b_2} , it is parallel to the vector \overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}

Now,  

\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}

\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 8 & - 16 & 7 \\ 3 & 8 & - 5\end{vmatrix}

24 \hat{i} + 61 \hat{j} + 112 \hat{k}

The direction ratios of the required line are proportional to 24, 61, 112. 

The equation of the required line passing through the point (1, 2,-4) and having direction ratios proportional to 24, 61, 112 is,

=> \frac{x - 1}{24} = \frac{y - 2}{61} = \frac{z + 4}{112}

Question 19. Show that the lines \frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1}  and \frac{x}{1} = \frac{y}{2} = \frac{z}{3}  are perpendicular to each other.

Solution:

The direction ratios of the line \frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1}  are proportional to 7, -5, 1 respectively.  

And the direction ratios of the line \frac{x}{1} = \frac{y}{2} = \frac{z}{3}  are proportional to 1, 2, 3 respectively.  

Let,

\overrightarrow{b_1} = 7 \hat{i} - 5 \hat{j} + \hat{k}
\overrightarrow{b_2} = \hat{i}  + 2 \hat{j}  + 3 \hat{k}

Now, 

\overrightarrow{b_1} . \overrightarrow{b_2} = \left( 7 \hat{i} - 5 \hat{j} + \hat{k} \right) . \left( \hat{i} + 2 \hat{j}  + 3 \hat{k}  \right)

= 7 – 10 + 3

= 0

So, \overrightarrow{b_1} \perp \overrightarrow{b_2}

Therefore, the given lines are perpendicular to each other.

Hence proved.

Question 20. 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. 

Solution:

The equation of the line 6x − 2 = 3y + 1 = 2z − 2 can be re-written as

\frac{x - \frac{1}{3}}{\frac{1}{6}} = \frac{y + \frac{1}{3}}{\frac{1}{3}} = \frac{z - 1}{\frac{1}{2}}

=> \frac{x - \frac{1}{3}}{1} = \frac{y + \frac{1}{3}}{2} = \frac{z - 1}{3}

Since the required line is parallel to the given line, the direction ratios of the required 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,

=> \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) + \lambda\left( \hat{i}  + 2 \hat{j} + 3 \hat{k}  \right)

Question 21. If the lines \frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2}  and \frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}  are perpendicular, find the value of λ.

Solution:

The equations of the given lines are,

\frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2}
\frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}

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

=> -3 (3λ) + 2λ (1) + 2 (-5) = 0

=> -9λ + 2λ – 10 = 0

=> -7λ = 10

=> λ = -10/7

Therefore, the value of λ is -10/7.

Question 22. 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. 

Solution:

The direction ratios of AB and CD are proportional to 3, 3, 4 and 6, 6, 8, respectively.

Let θ be the angle between AB and CD. Then,

\cos \theta = \frac{3 \times 6 + 3 \times 6 + 4 \times 8}{\sqrt{3^2 + 3^2 + 4^2} \sqrt{6^2 + 6^2 + 8^2}}

\frac{68}{\sqrt{34} \sqrt{136}}

= 1

Now cos θ = 1

=> θ = 0°

Therefore, the angle between AB and CD is 0°.

Question 23. Find the value of λ so that the following lines are perpendicular to each other. 

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

Solution:

The equation of the given line \frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}  can be re-written as,

\frac{x - 5}{5\lambda + 2} = \frac{y - 2}{- 5} = \frac{z - 1}{1}

The equation of the given line \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}  can be re-written as, 

\frac{x}{1} = \frac{y + \frac{1}{2}}{2\lambda} = \frac{z - 1}{3}

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

=> (5λ + 2) (1) – 5 (2λ) + 1 (3) = 0

=> 5λ + 2 – 10λ + 3 = 0

=> -5λ = -5

=> λ = 1

Therefore, the value of λ is 1.

Question 24. Find the direction cosines of the line \frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6} . Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.  

Solution:

The equation of the given line is,

\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}

The given equation can be re-written as

\frac{x + 2}{2} = \frac{y - \frac{7}{2}}{3} = \frac{z - 5}{- 6}

This line passes through the point (-2, 7/2, 5) and has direction ratios proportional to 2, 3, −6. 

So, its direction cosines are \left(\frac{2}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}, \frac{3}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}, \frac{- 6}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}\right)  

Or, \left(\frac{2}{7}, \frac{3}{7}, \frac{- 6}{7}\right)

The required line passes through the point having position vector \overrightarrow{a} = - \hat{i} + 2 \hat{j} + 3 \hat{k} .

And also it is parallel to the vector \overrightarrow{b} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} .

So, its vector equation is,

=> \overrightarrow{r} = \left( - \hat{i}  + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right)

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