第 12 课 RD Sharma 解决方案 – 第 28 章空间直线 – 练习 28.2 |设置 1

问题 1. 证明方向余弦为 12/13、-3/13、-4/13 的三条线； 4/13、12/13、3/13； 3/13, – 4/13, 12/13 相互垂直。

解决方案：

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.

问题 2. 证明通过点 (1, -1, 2) 和 (3, 4, -2) 的线垂直于通过点 (0, 3, 2) 和 (3, 5, 6) 的线。

解决方案：

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.

问题 3. 证明通过点 (4, 7, 8) 和 (2, 3, 4) 的线平行于通过点 (-1, -2, 1) 和 (1, 2, 5) 的线）。

解决方案：

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.

问题 4. 求通过点 (-2, 4, -5) 并平行于由下式给出的线的直线的笛卡尔方程$\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6}$ .

解决方案：

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,

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

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

问题 5. 显示线条$\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1}$和$\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$是相互垂直的。

解决方案：

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.

问题 6. 证明连接原点和点 (2, 1, 1) 的线垂直于由点 (3, 5, -1) 和 (4, 3, -1) 确定的线。

解决方案：

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.

问题 7. 求一条平行于 x 轴并通过原点的直线方程。

解决方案：

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

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

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

问题 8. 求以下一对直线之间的夹角：

（一世） $\vec{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right)$和$\vec{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)$

解决方案：

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°.

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

解决方案：

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).

㈢ $\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right)$和$\overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}$

解决方案：

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.

问题 9. 求下列一对直线之间的夹角：

（一世） $\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4}$和$\frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}$

解决方案：

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).

(二) $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3}$和$\frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}$

解决方案：

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$\cos^{- 1} \left( \frac{10}{9\sqrt{22}} \right)$.

㈢ $\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3}$和$\frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}$

解决方案：

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).

(四) $\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5$和$\frac{x + 1}{1} = \frac{2y - 3}{3} = \frac{z - 5}{2}$

解决方案：

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.

(五) $\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1}$和$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}$

解决方案：

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).

(六) $\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3}$和$\frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}$

解决方案：

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.

问题 10. 求方向比与以下成比例的线对之间的角度：

(i) 5、-12、13 和 -3、4、5

解决方案：

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 和 4、1、8

解决方案：

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 和 -2、2、1

解决方案：

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 和 b-c、c-a、a-b

解决方案：

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.

问题 11. 求两条线之间的夹角，其中一条的方向比为 2, 2, 1，而另一条是通过连接点 (3, 1, 4) 和 (7, 2, 12) 得到的。

解决方案：

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).

问题 12. 求通过点 (1, 2, -4) 并平行于直线的直线方程$\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{z + 1}{3}$ .

解决方案：

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

$\frac{x - 1}{4} = \frac{y - 2}{2} = \frac{z - \left( - 4 \right)}{3}$

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