Class 12 RD Sharma Solutions - Chapter 28 The Straight Line in Space - Exercise 28.2 | Set 1

Chapter 28 of RD Sharma's Class 12 Mathematics textbook focuses on the geometry of straight lines in three-dimensional space. Exercise 28.2 Set 1 specifically deals with various aspects of lines in space, including their equations, direction cosines, and relationships between lines. This set challenges students to apply their understanding of vector algebra and analytical geometry to solve problems involving lines in 3D coordinate systems.

Important Formulas for Straight Lines in Space

Vector equation: r = a + λb

Parametric equations: x = x₀ + at, y = y₀ + bt, z = z₀ + ct

Cartesian equation: (x - x₀)/l = (y - y₀)/m = (z - z₀)/n

Direction cosines: cos α = l/√(l²+m²+n²), cos β = m/√(l²+m²+n²), cos γ = n/√(l²+m²+n²)

Angle between two lines: cos θ = |l₁l₂ + m₁m₂ + n₁n₂| / √[(l₁²+m₁²+n₁²)(l₂²+m₂²+n₂²)]

Condition for parallel lines: l₁/l₂ = m₁/m₂ = n₁/n₂

Condition for perpendicular lines: l₁l₂ + m₁m₂ + n₁n₂ = 0

Distance of a point (x₀, y₀, z₀) from line: d = |[(x-x₀), (y-y₀), (z-z₀)] × (a, b, c)| / √(a²+b²+c²)

Distance between skew lines: d = |[(a₂-a₁) · (b₁×b₂)]| / |b₁×b₂|

Class 12 RD Sharma Mathematics Solutions- Exercise 28.2 | Set 1

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, m₃ = -4/13, n₃ = 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}

Practice Questions on Straight Line in Space

Question 1. Find the vector and cartesian equations of the line passing through the points (1, 2, 3) and (4, -1, 5).

Question 2. Write the vector equation of the line passing through the point (2, -1, 3) and parallel to the vector <1, 2, -1>.

Question 3. Find the cartesian equations of the line passing through (0, 1, -2) with direction ratios 2, 3, -1.

Question 4. Determine if the point (2, 3, 1) lies on the line (x-1)/2 = (y+1)/3 = (z-2)/4.

Question 5. Find the point where the line x = 1 + 2t, y = 2 - t, z = 3t intersects the xy-plane.

Question 6. Write the vector equation of the line of intersection of the planes x + y + z = 1 and 2x - y + z = 3.

Question 7. Find the cartesian equations of the line passing through the point (1, -1, 2) and perpendicular to the plane 2x + 3y - z = 5.

Question 8. Determine if the lines (x-2)/3 = (y+1)/4 = (z-3)/5 and (x-1)/2 = y/3 = (z+1)/4 are parallel, intersecting, or skew.

Question 9. Find the parametric equations of the line passing through the point (2, 3, -1) and making equal angles with the coordinate axes.

Question 10. Write the vector equation of the line passing through the point of intersection of the lines x + y = 2, z = 3 and x - y = 0, z = 1, and parallel to the line x = y = z.

Also Read,

• Straight Line in Space
• Class 12 RD Sharma Mathematics Solutions- Chapter 28 The Straight Line in Space - Exercise 28.3
• Class 12 RD Sharma Mathematics Solutions- Chapter 28 The Straight Line in Space - Exercise 28.5
• Class 12 RD Sharma Mathematics Solutions- Chapter 28 The Straight Line in Space - Exercise 28.1| Set 1

Conclusion

Exercise 28.2 Set 1 in RD Sharma's chapter on The Straight Line in Space provides students with a comprehensive set of problems to master the concepts of three-dimensional geometry as applied to straight lines. This exercise set challenges students to work with various forms of line equations, analyze relationships between lines, and solve complex spatial problems. By tackling these problems, students develop crucial skills in visualizing and manipulating geometric entities in 3D space, a fundamental ability for many advanced mathematical and scientific disciplines. The problems likely range from basic calculations involving direction cosines and line equations to more complex scenarios requiring the analysis of line intersections, angles between lines, and distances in space.