Class 12 RD Sharma Solutions - Chapter 28 The Straight Line in Space - Exercise 28.1 | Set 2

In Class 12 Mathematics, Chapter 28 focuses on understanding the concept of a straight line in three-dimensional space. This topic is crucial for analyzing geometric relationships and solving problems involving the lines and planes in 3D. Exercise 28.1 | Set 2 provides the set of problems designed to deepen your understanding of these concepts.

The Straight Line in Space

A straight line in space can be defined using various methods such as parametric equations, symmetric equations, and vector equations. The line's direction can be described by the direction vector while its position can be determined using the point through which it passes. These concepts are fundamental in spatial geometry helping us visualize and solve problems related to the lines intersecting or being parallel to the planes and other lines.

Question 11. Find the direction cosines of the line\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3} . Also, reduce it to vector form

Solution:

Given:

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

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

x = -2λ + 4, y = 6λ, z = -3λ + 1

So,

x\hat{i}+y\hat{j}+z\hat{k}=(-2λ+4)\hat{i}+(6λ)\hat{j}+(-3λ+1)\hat{k}\\ \vec{r}=(4\hat{i}+\hat{k})+λ(-2\hat{i}+6\hat{j}-3\hat{k})

Direction ratios of the line are = -2, 6, -3

Direction cosines of the lines are,

\frac{a}{\sqrt{a^2+b^2+c^2}},\ \frac{b}{\sqrt{a^2+b^2+c^2}},\ \frac{c}{\sqrt{a^2+b^2+c^2}}\\ \frac{-2}{\sqrt{(-2)^2+(6)^2+(-3)^2}},\ \frac{6}{\sqrt{(-2)^2+6^2+(-3)^2}},\ \frac{-3}{\sqrt{(-2)^2+6^2+(-3)^2}}\\ \frac{-2}{7},\ \frac{6}{7},\ \frac{-3}{7}

Question 12. The Cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

Solution:

x = ay + b

z = cy + d

\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}=\lambda

So, DR's of line are (a, 1, c)

From above equation, we can write

x = aλ + b

y = λ

z = cλ + d

So vector equation of line is

x\hat{i}+y\hat{j}+z\hat{k}=(b\hat{i}+d\hat{k})+λ(a\hat{i}+\hat{j}+c\hat{k})

Question 13. Find the vector equation of a line passing through the point with position vector\hat{i}-2\hat{j}-3\hat{k} and parallel to the line joining the points with the position vector\hat{i}-\hat{j}+4\hat{k} and2\hat{i}+\hat{j}+2\hat{k} . Also, find the Cartesian equivalent of this equation.

Solution:

We know that, equation of a line passing through\vec{a} and parallel to vector\vec{b} is

\vec{r}=\vec{a}+λ\vec{b} ....... (i)

Here,

\vec{a}=\hat{i}-2\hat{j}-3\hat{k}

and, \vec{b} = line joining(\hat{i}-\hat{j}+4\hat{k}) and(2\hat{i}+\hat{j}+2\hat{k})

=(2\hat{i}+\hat{j}+2\hat{k})-(\hat{i}-\hat{j}+4\hat{k})\\ =2\hat{i}-\hat{i}+\hat{j}+\hat{j}+2\hat{k}-4\hat{k}\\ =\hat{i}+2\hat{j}-2\hat{k}

Equation of the line is

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

For Cartesian form of equation putx\hat{i}+y\hat{j}+z\hat{k}

x\hat{i}+y\hat{j}+z\hat{k}=(1+\lambda)+\hat{i}(-2+2\lambda)\hat{j}+(-3-2\lambda)\hat{k}

Equating coefficients of\hat{i},\ \hat{j},\ \hat{k}

x = 1 + λ, y = -2 + 2λ, z = -3 - 2λ

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

Question 14. Find the points on the line\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2} at a distance of 5 units from the points P(1, 3, 3).

Solution:

Given, line is\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda

General points Q on line is (3λ - 2, 2λ -1), 2λ + 3)

Distance of points P from Q =\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

PQ =\sqrt{(3λ-2-1)^2+(2λ-1-3)^2+(2λ+3-3)^2}

(5)² = (3λ -3)² + (2λ - 4) + (2λ)²

25 = 9λ² + 9 - 18λ + 4λ² + 16 - 16λ + 4λ²

17λ² - 34λ = 0

17λ (λ - 2) = 0

λ = 0 or 2

So, points on the line are (3(0) - 2, 2(0) - 1, 2(0) + 3)

(3(2) - 2, 2(2) - 1, 2(2) + 3)

= (-2, -1, 3), (4, 3, 7)

Question 15. Show that the points whose position vectors are-2\hat{i}+3\hat{j},\ \hat{i} + 2\hat{j}+3\hat{k} and7\hat{i}-\hat{k} are collinear.

Solution:

Let the given points are A,B.C with position vectors\vec{a},\ \vec{b},\ \vec{c} respectively.

\vec{a}=2\hat{i}+3\hat{j},\ \vec{b}=\hat{i}+2\hat{j}+3\hat{k},\ \vec{c}=7\hat{i}-\hat{k}

We know that, equation of a line passing through\vec{a} and\vec{b} are

\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\\ =(-2\hat{i}+3\hat{j})+λ((\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j}))\\ =(-2\hat{i}+3\hat{j})+\lambda(\hat{i}+2\hat{j}+3\hat{k}+2\hat{i}-3\hat{j})\\ \vec{r}=(-2\hat{i}+3\hat{j})+\lambda(3\hat{i}-\hat{j}+3\hat{k})\ \ \ \ .....(i)

If A, B, C are collinear then\vec{c} must satisfy equation (i)

7\hat{i}-\hat{k}=(-2+3\lambda)\hat{i}+(3-\lambda)\hat{j}+(3\lambda)\hat{k}

Equation the coefficients of\vec{i},\ \vec{j},\ \vec{k}

-2 + 3 = 7 , λ = 3

3 - λ = 0 , λ = 3

3λ = -1 , λ =-\frac{1}{3}

Since, value of λ are not equal, so,

Given points are collinear.

Question 16. Find the Cartesian and vector equations of a line which passes through the points (1, 2, 3) and is parallel to the line\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}

Solution:

We know that, equation of a line passing through a point (x₁, y₁, z₁) and having direction ratios proportional to a, b, c is

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\ \ \ \ .....(i)

Here,

(x₁, y₁, z₁) = (1, 2, 3) and

Given line\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}

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

Its parallel to the required line, so

a = μ , b = 7μ, c =\frac{3}{2} μ

So, equation of required line using equation (i) is,

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

Multiplying the denominators by 2

\frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3}=λ

x = -2λ + 1, y = 14λ + 2, z = 3λ + 3

So, vector form of the equation of required line,

x\hat{i}+y\hat{j}+z\hat{k}=(-2λ+1)\hat{i}+(14λ+2)\hat{j}+(3λ+3)\hat{k}\\ \vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+λ(-2\hat{i}+14\hat{j}+3\hat{k})

Question 17. The Cartesian equations of a line are 3x + 1 = 6y - 2 = 1 - z. Find the fixed point through which it passes, its direction ratios, and also its vector equation.

Solution:

Given equation of line is,

3x + 1 = 6y -2 = 1 - z

Dividing all by 6

=\frac{3x+1}{6}=\frac{6y-2}{6}=\frac{1-z}{6}\\ =\frac{3x}{6}+\frac{1}{6}=\frac{6y}{6}-\frac{2}{6}=\frac{1}{6}-\frac{z}{6}\\ =\frac{1}{2}x+\frac{1}{6}=y-\frac{1}{3}=-\frac{z}{6}+\frac{1}{6}\\ =\frac{1}{2}\left(x+\frac{1}{3}\right)=1\left(y-\frac{1}{3}\right)=\frac{1}{6}(z-1)\\ =\frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}=λ

Comparing it with equation of line equation of line passing through (x,₁ y₁, z₁) and the direction ratios a, b, c,

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ (x_1,\ y_1,\ z_1)=\left(-\frac{1}{3},\ \frac{1}{3},\ 1\right)

a = 2, b = 1, -6

So, direction ratios of the line are -2, 1, -6

From equation (i)

x = \left(2λ-\frac{1}{3}\right),\ y=\left(λ+\frac{1}{3}\right),\ z=(-6λ+1)

So, vector equation of the given line is,

x\hat{i}+y\hat{j}+z\hat{k}=\left(2λ-\frac{1}{3}\right)\hat{i}+\left(λ+\frac{1}{3}\right)\hat{j}+(-6λ+1)\hat{k}\\ \vec{r}=\left(-\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k}\right)+λ(2\hat{i}+\hat{j}-6\hat{k})

Read More:

• Standard Form of a Straight Line
• How to find the Equation of a Straight Line?

Summary

Chapter 28 of RD Sharma Class 12 Solutions focuses on the straight line in three-dimensional space. Key topics covered include:

• Direction cosines and direction ratios
• Equations of lines in various forms (vector, parametric, symmetric)
• Angle between lines
• Shortest distance between skew lines
• Intersection of lines and planes
• Perpendicular lines and planes