RD Sharma Class 12 Exercise 27.5 Straight line in space Solutions Maths

RD Sharma Class 12 Exercise 27.5 Straight line in space Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Solutions Chapter27 Straight line in space - Other Exercise

Straight Line in Space Excercise: 27.5

Straight Line in Space exercise 27.5 question 1(i)

Answer:

d = 3 \sqrt[2]{30}

Hint: Using formula $a \frac{\left|

\begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}

\right|}{\sqrt{\left(a_{1} b-a_{2} b_{1}\right)^{2}+\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(a_{1} c_{2}-a_{2} c_{1}\right)^{2}}}$

Given:

\vec{γ} = (3 \hat{ı} + 8 \hat{ȷ} + 3 \hat{k}) + λ (3 \hat{ı} - \hat{ȷ} + \hat{k}) and (- 3 \hat{ı} - \hat{ȷ} + \hat{k}) + μ (7 \hat{ı} - 6 \hat{ȷ} + \hat{k})

Solution: From the given pair of equation
$

\begin{aligned} \frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \Rightarrow 1 \\ \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{- 4} \Rightarrow 2 \end{aligned}

$
$d=\frac{\left|

\begin{array}{ccc} 6 & 15 & - 3 \\ 3 & - 1 & 1 \\ - 3 & 2 & 4 \end{array}

\right|}{\sqrt{(6-3)^{2}+(-4-2)^{2}+(12+3)^{2}}}$
$

\begin{aligned} = | \frac{6 (- 4 - 2) - 15 (12 + 3) - 3 (6 - 3)}{\sqrt{9 + 36 + 225}} | \\ = \frac{- 36 - 225 - 9}{\sqrt{270}} \end{aligned}

$
$

\begin{aligned} = \frac{270}{\sqrt{270}} \\ = \sqrt{270} \\ = 3 \sqrt{30} \Rightarrow ANS \end{aligned}

Straight Line in Space exercise 27.5 question 1(ii)

Answer:

D = \frac{64}{\sqrt{62}} units

Hint: Using formula

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \overset{―}{b_{2}}) |}

Given:

\vec{γ} = (3 \hat{ı} + 5 \hat{ȷ} + 7 \hat{k}) + λ (\hat{ı} - 2 \hat{ȷ} + 7 \hat{k}) and \vec{γ} = (- \hat{ı} - \hat{ȷ} - \hat{k}) + μ (7 \hat{ı} - 6 \hat{ȷ} + \hat{k})

Solution:
$

\begin{aligned} \vec{a_{1}} = 3 \hat{ı} + 5 \hat{ȷ} + 7 \hat{k} \\ \vec{a_{2}} = - \hat{ı} - \hat{ȷ} - \hat{k} \\ \vec{b_{1}} = \hat{ı} - 2 \hat{ȷ} + 7 \hat{k} \\ \vec{b_{2}} = 7 \hat{ı} - 6 \hat{ȷ} + \hat{k} \end{aligned}

$
The shortest distance between the lines is
$

\begin{matrix} d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \overset{―}{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |} \\ (\vec{a_{2}} - \vec{a_{1}}) = (- \hat{ı} - \hat{ȷ} - \hat{k}) - (3 \hat{ı} + 5 \hat{ȷ} + 7 \hat{k}) \\ = - 4 \hat{ı} - 6 \hat{ȷ} - 8 \hat{k} \end{matrix}

$
$

\begin{aligned} (\vec{b_{1}} \times \vec{b_{2}}) & = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 1 & - 2 & 7 \\ 7 & - 6 & 1 \end{array} | \\ (\vec{b_{1}} \times \vec{b_{2}}) & = (- 2 + 42) + [- (1 - 49) \hat{j}] + (- 6 + 14 \hat{k}) \\ = 40 \hat{ı} + 48 \hat{ȷ} + 8 \hat{k} \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | & = \sqrt{(40)^{2} + (48)^{2} + (8)^{2}} \\ = 8 \sqrt{62} \end{aligned}

$
$

\begin{aligned} | (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | & = | (- 4 \hat{ı} - 6 \hat{ȷ} - 8 \hat{k}) (40 \hat{ı} + 48 \hat{ȷ} + 8 \hat{k}) | \Rightarrow \\ | (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | & = | - 512 | \\ = 512 \end{aligned}

$
Putting the values in the above mentioned formula,
$

\begin{aligned} d = \frac{512}{8 \sqrt{62}} \\ = \frac{64}{\sqrt{62}} \\ = \frac{64}{\sqrt{62}} units \end{aligned}

Straight Line in Space exercise 27.5 question 1(iii)

Answer: $d = \frac{1}{\sqrt{6}} units$
Hint: Using formula

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given: $\vec{γ} = (\hat{ı} + 2 \hat{ȷ} + 3 \hat{k}) + λ (2 \hat{ı} + 3 \hat{ȷ} + 4 \hat{k}) and \vec{γ} = (2 \hat{ı} + 4 \hat{ȷ} + 5 \hat{k}) + μ (3 \hat{ı} + 4 \hat{ȷ} + 5 \hat{k})$
Solution:
By the given pairs of equation,
$

\begin{aligned} \vec{a_{1}} = \hat{ı} + 2 \hat{ȷ} + 3 \hat{k} \\ \vec{a_{2}} = 2 \hat{ı} + 4 \hat{j} + 5 \hat{k} \\ \vec{b_{1}} = 2 \hat{ı} + 3 \hat{ȷ} + 5 \hat{k} \\ \vec{b_{2}} = 3 \hat{ı} + 4 \hat{ȷ} + 5 \hat{k} \end{aligned}

$
The shortest distance between the lines is,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = (2 \hat{ı} + 4 \hat{ȷ} + 5 \hat{k}) - (\hat{ı} + 2 \hat{ȷ} + 3 \hat{k}) \\ (\vec{a_{2}} - \vec{a_{1}}) = \hat{ı} + 2 \hat{ȷ} + 2 \hat{k} \end{aligned}

$
$

\begin{aligned} (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array} | \\ (\vec{b_{1}} \times \vec{b_{2}}) = - \hat{ı} + 2 \hat{ȷ} - \hat{k} \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(- 1)^{2} + (2)^{2} + (- 1)^{2}} \\ | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{6} \\ | (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | = | (\hat{ı} + 2 \hat{ȷ} + 2 \hat{k}) (- \hat{ı} + 22 \hat{ȷ} - \hat{k}) | \Rightarrow 1 \end{aligned}

$
Putting three values in the expression,
$

\begin{aligned} d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |} \\ d = \frac{1}{\sqrt{6}} units \end{aligned}

Straight Line in Space exercise 27.5 question 1(iv)

Answer: $d = \frac{8}{\sqrt{29}} units$
Hint: Using formula

= \frac{| (\vec{c} - \vec{a}) (\vec{b} \times \vec{d}) |}{| (\vec{b} \times \vec{d}) |}

and convert the given equation in normal forms.

Given: $\vec{γ} = (1 - t) \hat{ı} + (t - 2) \hat{ȷ} + (3 - t) \hat{k} and \vec{γ} = (s +) \hat{ı} + (2 s - 1) \hat{ȷ} - (2 s + 1) \hat{k}$
Solution:
$ $\begin{array}{ll} \vec{γ} = (\hat{ı} + 2 \hat{ȷ} + 3 \hat{k}) + t (- \hat{ı} + \hat{ȷ} - 2 \hat{k}) & [\vec{γ} = \vec{a} + λ \vec{b}] \\ \vec{γ} = (\hat{ı} - \hat{ȷ} - \hat{k}) + s (\hat{ı} + 2 \hat{ȷ} - 2 \hat{k}) & [\vec{γ} = \vec{c} + λ \vec{d}] \end{array}$ $
$\vec{b} \times \vec{d}=\left| $\begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ - 1 & 1 & - 2 \\ 1 & 2 & - 2 \end{array}$ \right|$
$= \hat{ı} [1 (- 2) - (- 2) (2)] - \hat{ȷ} [(- 1) (- 2) - (1) (- 2)] + \hat{k} [(- 1) (2) - (1) (1)]$
$ $\begin{aligned} \vec{b} \times \vec{d} = 2 \hat{ı} - 4 \hat{ȷ} - 3 \hat{k} \\ | \vec{b} \times \vec{d} | = \sqrt{(2)^{2} + (- 4)^{2} + (- 3)^{2}} \Rightarrow \sqrt{9 + 16 + 4} \Rightarrow \sqrt{29} \end{aligned}$ $
$ $\begin{aligned} (\vec{c} - \vec{a}) & = (\hat{ȷ} - 4 \hat{k}) \\ d & = \frac{| (\hat{ȷ} - 4 \hat{k}) (2 \hat{ı} - 4 \hat{ȷ} - 3 \hat{k}) |}{| \sqrt{29} |} \Rightarrow \frac{| 0 - 4 + 12 |}{| \sqrt{29} |} \Rightarrow \frac{8}{\sqrt{29}} \end{aligned}$ $

Straight Line in Space exercise 27.5 question 1(v)

Answer: $d = \frac{5}{2} \sqrt{2} units$
Hint: Convert the given equation in the normal form

\vec{γ} = \vec{a} + λ \vec{b}

\begin{aligned} Let \vec{γ} & = a_{1} + λ b_{1} \\ \vec{γ} & = a_{2} + μ b_{2} \end{aligned}

$
Given: $\vec{γ} = (λ - 1) \hat{ı} + (λ + 1) \hat{ȷ} - (1 + λ) \hat{k} and \vec{γ} = (1 - μ) \hat{ı} + (2 μ - 1) \hat{ȷ} + (μ + 2) \hat{k}$
Solution:
$ $\begin{aligned} \vec{γ} = (- \hat{ı} + \hat{ȷ} - \hat{k}) + λ (\hat{ı} + \hat{ȷ} - \hat{k}) and \\ \vec{γ} = (\hat{ı} - \hat{ȷ} + 2 \hat{k}) + μ (- \hat{ı} + 2 \hat{ȷ} + \hat{k}) \\ D = \frac{| (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) |}{| (b_{1} \times b_{2}) |} \end{aligned}$ $
Here
$

\begin{array}{ll} \vec{a_{1}} = - \hat{ı} + \hat{ȷ} - \hat{k} & \vec{b_{1}} = \hat{ı} + \hat{ȷ} - \hat{k} \\ \vec{a_{2}} = \hat{ı} - \hat{ȷ} + 2 \hat{k} & \vec{b_{2}} = - \hat{ı} \\ (\vec{a_{2}} - \vec{a_{1}}) = 2 \hat{i} - 2 \hat{ȷ} + 3 \hat{k} \end{array}

$
$\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|

\begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 1 & 1 & - 1 \\ - 1 & 2 & 1 \end{array}

\right|$
$

\begin{aligned} \vec{b_{1}} \times \vec{b_{2}} & = \hat{ı} (1 + 2) - \hat{ȷ} (- 1 - 1) + \hat{k} (2 + 1) \\ = 3 \hat{ı} + 3 \hat{k} \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(3)^{2} + (3)^{2}} \Rightarrow 3 \sqrt{2} \\ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = (2 \hat{ı} - 2 \hat{ȷ} + 3 \hat{k}) \cdot (3 \hat{ı} + 3 \hat{k}) \Rightarrow 6 + 9 \Rightarrow 15 \\ D = \frac{15}{3 \sqrt{2}} \Rightarrow \frac{15}{\sqrt{2}} \Rightarrow \frac{5}{2 \sqrt{2}} \end{aligned}

Straight Line in Space exercise 27.5 question 1(vi)

Answer: $d = \frac{3}{\sqrt{2}} units$
Hint: Using the expression

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given: $\vec{γ} = (2 \hat{ı} - \hat{ȷ} - \hat{k}) + λ (2 \hat{ı} - 5 \hat{ȷ} + 2 \hat{k}) and \vec{γ} = (\hat{ı} + 2 \hat{j} + \hat{k}) + μ (\hat{ı} - \hat{ȷ} + \hat{k})$
Solution:
$ $\begin{array}{ll} \vec{a_{1}} = 2 \hat{ı} - \hat{ȷ} - \hat{k} & \vec{b_{1}} = 2 \hat{ı} - 5 \hat{ȷ} + 2 \hat{k} \\ \vec{a_{2}} = \hat{ı} + 2 \hat{ȷ} + \hat{k} & \vec{b_{2}} = \hat{ı} - \hat{ȷ} + \hat{k} \end{array}$ $
The shortest distance between the lines is,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = (\hat{ı} + 2 \hat{ȷ} + \hat{k}) - (2 \hat{ı} - \hat{ȷ} - \hat{k}) \\ (\vec{a_{2}} - \vec{a_{1}}) = - \hat{ı} + 3 \hat{ȷ} + 2 \hat{k} \end{aligned}

$
$\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|

\begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 2 & - 5 & 2 \\ 1 & - 1 & 1 \end{array}

\right|$
$

\begin{aligned} (\vec{b_{1}} \times \vec{b_{2}}) = (- 5 + 2) \hat{ı} - (2 - 2) \hat{ȷ} + (- 2 + 5) \hat{k} \\ | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(- 3)^{2} + (0)^{2} + (3)^{2}} \\ | \vec{b_{1}} \times \vec{b_{2}} | = 3 \sqrt{2} \end{aligned}

| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | = | (- \hat{ı} + 3 \hat{ȷ} + 2 \hat{k}) (- 3 \hat{ı} + 0 \hat{ȷ} + 3 \hat{k}) | \Rightarrow 9

Putting three values in the above mentioned equation,
$

\begin{aligned} d = \frac{9}{3 \sqrt{2}} \\ d = \frac{3}{\sqrt{2}} units \end{aligned}

Straight Line in Space exercise 27.5 question 1(vii)

Answer: $d = \frac{10}{\sqrt{59}} units$
Hint: Using the expression

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given: $\vec{γ} = (\hat{ı} + \hat{ȷ}) + λ (2 \hat{ı} - \hat{ȷ} + \hat{k}) and \vec{γ} = (2 \hat{ı} + \hat{ȷ} - \hat{k}) + μ (3 \hat{ı} - 5 \hat{ȷ} + 2 \hat{k})$
Solution: We know that,
$

\begin{array}{ll} \vec{a_{1}} = \hat{ı} + \hat{ȷ} & \vec{b_{1}} = 2 \hat{ı} - \hat{ȷ} + \hat{k} \\ \vec{a_{2}} = 2 \hat{ı} + \hat{ȷ} - \hat{k} & \vec{b_{2}} = 3 \hat{ı} - 5 \hat{ȷ} + 2 \hat{k} \end{array}

$
It can be written as,
$

\begin{aligned} (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 2 & - 1 & 1 \\ 3 & - 5 & 2 \end{array} | \\ (\vec{b_{1}} \times \vec{b_{2}}) = (- 2 + 5) \hat{ı} - (4 - 3) \hat{ȷ} + (- 10 + 3) \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = 3 \hat{ı} - \hat{ȷ} - 7 \hat{k} \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(3)^{2} + (- 1)^{2} + (- 7)^{2}} \Rightarrow \sqrt{9 + 1 + 49} \Rightarrow \sqrt{59} \\ (\vec{a_{2}} - \vec{a_{1}}) = (2 - 1) \hat{ı} - (1 - 1) \hat{ȷ} + (- 1 - 0) \hat{k} = \hat{ı} + 0 \hat{ȷ} - \hat{k} \end{aligned}

$
We can write as,
$

\begin{aligned} (\vec{b_{1}} \times \vec{b_{2}}) \cdot (\vec{a_{2}} - \vec{a_{1}}) & = (3 \hat{ı} - \hat{ȷ} - 7 \hat{k}) \cdot (\hat{ı} + 0 \hat{ȷ} - \hat{k}) \\ = (3 \times 1) + [(- 1) \times 0] + ([(- 7) \times (- 1) \\ = 3 + 0 + 7 \Rightarrow 10 \end{aligned}

$
The shortest distance between the given lines is,

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |} \Rightarrow \frac{10}{\sqrt{59}} units

Straight Line in Space exercise 27.5 question 1(viii)

Answer: $d = 14 units$
Hint: Using the expression to find

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given: $\vec{γ} = (8 + 3 λ) \hat{ı} - (9 + 16 λ) \hat{ȷ} + (10 + 7 λ) \hat{k} and \vec{γ} = (15 \hat{ı} + 29 \hat{ȷ} + 5 \hat{k}) +$ $μ (3 \hat{ı} + 8 \hat{ȷ} - 2 \hat{k})$

Solution: We know that,
$

\begin{array}{ll} \vec{a_{1}} = 8 \hat{ı} - 9 \hat{ȷ} + 10 \hat{k} & \vec{b_{1}} = 3 \hat{ı} - 16 \hat{ȷ} + 7 \hat{k} \\ \vec{a_{2}} = 15 \hat{ı} + 29 \hat{ȷ} + 5 \hat{k} & \vec{b_{2}} = 3 \hat{ı} + 8 \hat{ȷ} - 5 \hat{k} \end{array}

$
The shortest distance between the given lines is,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = (15 \hat{ı} + 29 \hat{ȷ} + 5 \hat{k}) - (8 \hat{ı} - 9 \hat{ȷ} - 10 \hat{k}) \\ (\vec{a_{2}} - \vec{a_{1}}) = 7 \hat{ı} + 38 \hat{ȷ} - 5 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 3 & - 16 & 7 \\ 3 & - 8 & - 5 \end{array} | \end{aligned}

$
$

\begin{aligned} (\vec{b_{1}} \times \vec{b_{2}}) = (80 - 56) \hat{ı} - (- 15 - 21) \hat{ȷ} + (24 + 48) \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = 24 \hat{ı} + 36 \hat{ȷ} + 72 \hat{k} \Rightarrow 4 (6 \hat{ı} + 8 \hat{ȷ} + 18 \hat{k}) \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | = 4 \sqrt{(6)^{2} + (9)^{2} + (18)^{2}} \Rightarrow 4 \sqrt{441} \Rightarrow 4 \times 21 \Rightarrow 84 \\ | (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | = | 4 (7 \hat{ı} + 38 \hat{ȷ} - 5 \hat{k}) (6 \hat{ı} + 9 \hat{ȷ} + 18 \hat{k}) | \Rightarrow 4 (42 + 342 - 90) \Rightarrow 1176 \end{aligned}

$
Putting three values in the above mentioned equation,
$

\begin{aligned} d = \frac{1176}{84} \\ d = 14 units \end{aligned}

Straight Line in Space exercise 27.5 question 2(ii)

Answer: $d = \frac{1}{\sqrt{6}} units$
Hint: Using the expression to find

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given: $: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \to (1) and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{4} \to (2)$
Solution: Since the given line (1) passes through the point

(1, 2, 3)

and has direction ratios proportional to

(2, 3, 4)

its vector equation is

\vec{γ} = \vec{a_{1}} + λ \vec{b_{1}}

Here,
$

\begin{aligned} \vec{a_{1}} = \hat{ı} + 2 \hat{ȷ} + 3 \hat{k} \\ \vec{b_{1}} = 2 \hat{ı} + 3 \hat{ȷ} + 4 \hat{k} \end{aligned}

$
Also line (2) passes through the point

(2, 3, 5)

and has direction ratios proportional to

(3, 4, 5)

its vector equation is

\vec{γ} = \vec{a_{2}} + μ \vec{b_{2}}

Here,
$

\begin{aligned} \vec{a_{2}} = 2 \hat{ı} + 3 \hat{ȷ} + 5 \hat{k} \\ \vec{b_{2}} = 3 \hat{ı} + 4 \hat{ȷ} + 5 \hat{k} \end{aligned}

$
Now,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) & = \hat{ı} + \hat{j} + 2 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) & = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array} | \\ = - \hat{ı} + 2 \hat{ȷ} - \hat{k} \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(- 1)^{2} + (2)^{2} + (- 1)^{2}} \Rightarrow \sqrt{6} and \\ (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) = (\hat{ı} + \hat{ȷ} + 2 \hat{k}) (- \hat{ı} + 2 \hat{ȷ} - \hat{k}) \Rightarrow - 1 + 2 - 2 \Rightarrow - 1 \end{aligned}

$
The shortest distance between the given lines is,

d = | \frac{- 1}{\sqrt{6}} | \Rightarrow \frac{1}{\sqrt{6}} units

Straight Line in Space exercise 27.5 question 2(ii)

Answer: $d = \frac{3}{\sqrt{59}} units$
Hint: Consider the given equation as (1) and (2) and using the expression

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given: $: \frac{x - 1}{2} = \frac{y + 1}{3} = z \to (1) and \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2 \to (2)$
Solution:
Since line (1) passes through the point

(1, - 1, 0)

and has direction ratios proportional to

(2, 3, 1)

its vector equation is

\vec{γ} = \vec{a_{1}} + λ \vec{b_{1}}

Here,
$

\begin{aligned} \vec{a_{1}} = \hat{ı} - \hat{ȷ} + 0 \hat{k} \\ \vec{b_{1}} = 2 \hat{ı} + 3 \hat{ȷ} + \hat{k} \end{aligned}

$
Also line (2) passes through the point

(- 1, 2, 2)

and has direction ratios proportional to

(3, 1, 0)

its vector equation is

\vec{γ} = \vec{a_{2}} + μ \vec{b_{2}}

Here,
$

\begin{aligned} \vec{a_{2}} = - \hat{ı} + 2 \hat{ȷ} + 2 \hat{k} \\ \vec{b_{2}} = 3 \hat{ı} + \hat{ȷ} + 0 \hat{k} \end{aligned}

$
Now,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) & = - 2 \hat{ı} + 2 \hat{ȷ} + 2 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) & = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 2 & 3 & 1 \\ 3 & 1 & 0 \end{array} | \\ = - \hat{ı} + 3 \hat{ȷ} - 7 \hat{k} \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(- 1)^{2} + (3)^{2} + (- 7)^{2}} \Rightarrow \sqrt{1 + 9 + 49} \Rightarrow \sqrt{59} and \\ (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) = (- 2 \hat{ı} + 3 \hat{ȷ} + 2 \hat{k}) (- \hat{ı} + 3 \hat{ȷ} - 7 \hat{k}) \Rightarrow 2 + 9 - 14 \Rightarrow - 3 \end{aligned}

$
The shortest distance between the lines

\vec{γ} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{V} = \vec{a_{2}} + μ \vec{b_{2}}

is given by

d = | \frac{- 3}{\sqrt{59}} | \Rightarrow \frac{3}{\sqrt{59}} units

Straight Line in Space exercise 27.5 question 2(iii)

Answer:

d = \frac{8}{\sqrt{29}} units

Hint: Consider the given equation as (1) and (2)
Given:

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

Solution: Since line (1) passes through the point

(1, - 2, 3)

and has direction ratios proportional to

(- 1, 1, - 2)

its vector equation is

\vec{γ} = \vec{a_{1}} + λ \vec{b_{1}}

Here,
$

\begin{aligned} \vec{a_{1}} = \hat{ı} - 2 \hat{ȷ} + 3 \hat{k} \\ \vec{b_{1}} = - \hat{ı} + \hat{ȷ} - 2 \hat{k} \end{aligned}

$
Also line (2) passes through the point

(1, - 1, 1)

and has direction ratios proportional to

(1, 2, - 2)

its vector equation is

\vec{γ} = \vec{a_{2}} + μ \vec{b_{2}}

Here,
$

\begin{aligned} \vec{a_{2}} = \hat{ı} - \hat{ȷ} - \hat{k} \\ \vec{b_{2}} = \hat{ı} + 2 \hat{ȷ} - 2 \hat{k} \end{aligned}

$
Now,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) & = \hat{ȷ} - 2 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) & = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ - 1 & 1 & - 2 \\ 1 & 2 & - 2 \end{array} | \\ = 2 \hat{ı} - u \hat{ȷ} - 3 \hat{k} \end{aligned}

| \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(2)^{2} + (- 4)^{2} + (- 3)^{2}} \Rightarrow \sqrt{4 + 16 + 9} \Rightarrow \sqrt{29} and

The shortest distance between the lines
$

\begin{aligned} d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |} \\ d = | \frac{8}{\sqrt{29}} | \Rightarrow \frac{8}{\sqrt{29}} units \end{aligned}

Straight Line in Space exercise 27.5 question 2(iv)

Answer: $d = 2 \sqrt{29} units$
Hint: $Put λ = 0 and k = 0$
Given: $: \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \to (1) and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \to (2)$
Solution:

Now, let’s take a point on first line as

A (λ + 3, 2 λ + 5, λ + 7)

and let

B (7 k - 1, - 6 k - 1, k - 1)

be point on the second line.
The direction ratio of the line;

(7 k - λ - 4) \times 1 + (- 6 k + 2 λ - 6) \times (- 2) + (k - λ - 8) \times 1 = 0 Line (1) and

(7 k - λ - 4) \times 7 + (- 6 k + 2 λ - 6) \times (- 6) + (k - λ - 8) \times 1 = 0 Line(2)

Solving equation (1) and (2) we get
$

\begin{aligned} λ = 0 and k = 0 \\ A = (3, 5, 7) and B = (- 1, - 1, - 1) \\ A B = \sqrt{(3 + 1)^{2} + (5 + 1)^{2} + (7 + 1)^{2}} \Rightarrow \sqrt{16 + 32 + 64} \Rightarrow \sqrt{116} \Rightarrow 2 \sqrt{29} units \end{aligned}

Straight Line in Space exercise 27.5 question 3(i)

Answer: Given Lines are not interesting
Hint: Using the equation

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given: $\vec{γ} = (\hat{ı} - \hat{ȷ}) + λ (2 \hat{ı} + \hat{k}) and \vec{γ} = (2 \hat{ı} - \hat{ȷ}) + μ (\hat{ı} + \hat{ȷ} + \hat{k})$
Solution:
We know that,
$

\begin{array}{ll} \vec{a_{1}} = \hat{ı} - \hat{ȷ} + 0 \hat{k} & \vec{b_{1}} = 2 \hat{ı} + 0 \hat{ȷ} + \hat{k} \\ \vec{a_{2}} = 2 \hat{ı} - \hat{ȷ} & \vec{b_{2}} = \hat{ı} + \hat{ȷ} - \hat{k} \end{array}

$
The shortest distance between the lines is,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = (2 \hat{ı} - \hat{ȷ}) - (\hat{ı} - \hat{ȷ} + 0 \hat{k}) \\ (\vec{a_{2}} - \vec{a_{1}}) = \hat{ı} + 0 \hat{ȷ} + 0 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & - 1 \end{array} | \end{aligned}

$
$

\begin{aligned} (\vec{b_{1}} \times \vec{b_{2}}) = (0 - 1) \hat{ı} - (- 2 - 1) \hat{ȷ} + (2 - 0) \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = - \hat{ı} + 3 \hat{ȷ} + 2 \hat{k} \\ | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(- 1)^{2} + (3)^{2} + (2)^{2}} \Rightarrow \sqrt{14} \\ | (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | = | (\hat{ı} + 0 \hat{ȷ} + 0 \hat{k}) (- \hat{ı} + 3 \hat{ȷ} + 2 \hat{k}) | \Rightarrow 1 \end{aligned}

$
Putting three values in the above mentioned equation,
$

\begin{aligned} d = \frac{1}{\sqrt{14}} \\ d = \frac{1}{\sqrt{14}} units \end{aligned}

$
∴ Shortest distance d between the lines is not 0, so lines are not intersecting.

Straight Line in Space exercise 27.5 question 3(ii)

Answer: Given Lines are not interesting
Hint: Using the equation

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given: $\vec{γ} = (\hat{ı} + \hat{ȷ} - \hat{k}) + λ (3 \hat{ı} - \hat{ȷ}) and \vec{γ} = (4 \hat{ı} - \hat{k}) + μ (2 \hat{ı} + 3 \hat{k})$
Solution:
We know that,
$

\begin{array}{ll} \vec{a_{1}} = \hat{ı} + \hat{ȷ} - \hat{k} & \vec{b_{1}} = 3 \hat{ı} - 1 \hat{ȷ} + 0 \hat{k} \\ \vec{a_{2}} = 4 \hat{ı} + 0 \hat{ȷ} - \hat{k} & \vec{b_{2}} = 2 \hat{ı} + 0 \hat{ȷ} + 3 \hat{k} \end{array}

$
The shortest distance between the lines is,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = (4 \hat{ı} + 0 \hat{ȷ} - \hat{k}) - (\hat{ı} + \hat{ȷ} - \hat{k}) \\ (\vec{a_{2}} - \vec{a_{1}}) = 3 \hat{ı} - \hat{ȷ} + 0 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 3 & - 1 & 0 \\ 2 & 0 & 3 \end{array} | \end{aligned}

$
$

\begin{aligned} (\vec{b_{1}} \times \vec{b_{2}}) = (3 - 0) \hat{ı} - (9 - 0) \hat{ȷ} + (0 + 2) \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = - 3 \hat{ı} - 9 \hat{j} + 2 \hat{k} \\ | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(- 3)^{2} + (- 9)^{2} + (2)^{2}} \Rightarrow \sqrt{94} \\ | (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | = | (3 \hat{ı} - 1 \hat{ȷ} + 0 \hat{k}) (- 3 \hat{ı} - 9 \hat{ȷ} + 2 \hat{k}) | \Rightarrow 0 \end{aligned}

$
Putting three values in the above mentioned equation,
$

\begin{aligned} d = \frac{0}{\sqrt{94}} \\ d = 0 \end{aligned}

$
∴ Shortest distance d between the lines is 0, so lines are intersecting.

Straight Line in Space exercise 27.5 question 3(iii)

Answer: Given Lines are not interesting
Hint: Consider the given equation as (1) and (2)
Given: $: \frac{x - 1}{2} = \frac{y + 1}{3} = z \to (1) and \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2 \to (2)$
Solution: Since line (1) passes through the point

(1, - 1, 0)

and has direction ratios proportional to

(2, 3, 1)

its vector equation is

\vec{γ} = \vec{a_{1}} + λ \vec{b_{1}}

Here
$

\begin{aligned} \vec{a_{1}} = \hat{ı} - 1 \hat{ȷ} + 0 \hat{k} \\ \vec{b_{1}} = 2 \hat{i} + 3 \hat{j} + \hat{k} \end{aligned}

$
Also line (2) passes through the point

(- 1, 2, 2)

and has direction ratios proportional to

(5, 1, 0)

its vector equation is

\vec{γ} = \vec{a_{2}} + μ \vec{b_{2}}

Here,
$

\begin{aligned} \vec{a_{2}} = - \hat{ı} + 2 \hat{ȷ} + 2 \hat{k} \\ \vec{b_{2}} = 5 \hat{i} + 1 \hat{j} + 0 \hat{k} \end{aligned}

$
Now,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = - 2 \hat{ı} + 3 \hat{ȷ} + 2 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{lll} \hat{ı} & \hat{ȷ} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 0 \end{array} | \Rightarrow - \hat{ı} + 5 \hat{ȷ} - 13 \hat{k} \\ (\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) = (- 2 \hat{ı} + 3 \hat{ȷ} + 2 \hat{k}) \cdot (- \hat{ı} + 5 \hat{ȷ} - 13 \hat{k}) \Rightarrow 9 + 15 - 26 \Rightarrow 9 \end{aligned}

$
We observe,

(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) \neq 0

Thus, the given lines are not intersecting.

Straight Line in Space exercise 27.5 question 3(iv)

Answer: Given Lines are not interesting
Hint: Consider the given equation as (1) and (2)
Given:

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

Solution: Since line (1) passes through the point

(5, 7, - 3)

and has direction ratios proportional to

(4, - 5, - 5)

its vector equation is

\vec{γ} = \vec{a_{1}} + λ \vec{b_{1}}

Here,
$

\begin{aligned} \vec{a_{1}} = 5 \hat{ı} + 7 \hat{ȷ} - 3 \hat{k} \\ \vec{b_{1}} = 4 \hat{ı} - 5 \hat{ȷ} - 5 \hat{k} \end{aligned}

$
Also line (2) passes through the point

(8, 7, 5)

and has direction ratios proportional to

(7, 1, 3)

its vector equation is

\vec{γ} = \vec{a_{2}} + μ \vec{b_{2}}

Here,
$

\begin{aligned} \vec{a_{2}} = 8 \hat{ı} + 7 \hat{ȷ} + 5 \hat{k} \\ \vec{b_{2}} = 7 \hat{i} + 1 \hat{ȷ} + 3 \hat{k} \end{aligned}

$
Now,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = 3 \hat{ı} + 8 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 4 & - 5 & - 5 \\ 7 & 1 & 3 \end{array} | \Rightarrow - 10 \hat{ı} - 47 \hat{ȷ} + 39 \hat{k} \\ (\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) = (3 \hat{ı} + 8 \hat{k}) \cdot (- 10 \hat{ı} - 47 \hat{ȷ} + 39 \hat{k}) \Rightarrow - 30 + 312 \Rightarrow 282 \end{aligned}

$
We observe,

(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) \neq 0

Thus, the given lines are not intersecting.

Straight Line in Space exercise 27.5 question 4(i)

Answer: $d = \sqrt{26} units$
Hint: Let in

L 2

these is λ draw

∣∣

Line

Given: $\vec{γ} = (1 \hat{ı} + 2 \hat{ȷ} + 3 \hat{k}) + λ (1 \hat{ı} - 1 \hat{ȷ} + \hat{k}) and \vec{γ} = (2 \hat{ı} - \hat{ȷ} - \hat{k}) + μ (- \hat{ı} + \hat{ȷ} - \hat{k})$
Solution:
$ $\begin{aligned} L_{1} (1, 2, 3), \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{1} \\ L_{2} (2, - 1, - 1), \frac{x - 2}{1} = \frac{y + 1}{- 1} = \frac{z + 1}{1} = λ \end{aligned}$ $
Line (2) becomes

(λ + 2, - λ - 1, λ - 1)

\begin{aligned} = (λ + 2 - 1) \hat{ı} + (- λ - 1 - 2) \hat{ȷ} + (λ - 1 - 3) \hat{k} \\ = (λ + 1) \hat{ı} + (- λ - 3) \hat{ȷ} + (λ - 4) \hat{k} \end{aligned}

$
$

\begin{aligned} = (λ + 1) (1) + (- λ - 3) (- 1) + (λ - 4) (1) \\ = λ + 1 + λ + 3 + λ - 4 \\ = 3 λ = 0 \end{aligned}

∴ λ = 0

The shortest distance between the

∣∣

Line is

2, - 1, - 1 a n d (1, 2, 3)

Straight Line in Space exercise 27.5 question 4(ii)

Answer: $d = 0$
Hint:Using the equation

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Let the given lines L₁ and L₂
Given: $\vec{γ} = (\hat{ı} + \hat{ȷ}) + λ (2 \hat{ı} - \hat{ȷ} + \hat{k}) and \vec{γ} = (2 \hat{ı} + \hat{ȷ} - \hat{k}) + μ (4 \hat{ı} - 2 \hat{ȷ} + 2 \hat{k})$
Solution:
$ $\begin{aligned} L_{1} = a_{1} + λ b_{1} and L_{2} = a_{2} + μ b_{2} \\ (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 2 & - 1 & 1 \\ 4 & - 2 & 2 \end{array} | \Rightarrow \hat{ı} (0) - \hat{ȷ} (0) + \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = 0 \end{aligned}$ $
So the shortest distance is =0

Straight Line in Space exercise 27.5 question 5(i)

Answer: $d = 0$

Hint: $\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}}$

Given: $(0, 0, 0) and (1, 0, 2)$
Solution:
Applying the given point in the formula of the straight line
$

\begin{aligned} \frac{x - 0}{1 - 0} = \frac{y - 0}{0 - 0} = \frac{z - 0}{2 - 1} \\ \Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{2} \end{aligned}

$
Solution (ii)
Again applying the points

(1, 3, 0)

and

(0, 3, 0)

The equation of the line passing through the points

(1, 3, 0)

is,
$

\begin{aligned} \frac{x - 1}{0 - 1} = \frac{y - 3}{3 - 3} = \frac{z - 0}{2 - 1} \\ \Rightarrow \frac{x - 1}{- 1} = \frac{y - 3}{0} = \frac{z}{0} \end{aligned}

$
Since the first line passes through the points

(1, 3, 0)

and has the directions proportional to

(1, 0, 2)

its vector is,

\vec{γ} = \vec{a_{1}} + λ \vec{b_{1}} \to (1)

Hence,
$

\begin{aligned} \vec{a} = 0 \hat{ı} + 0 \hat{ȷ} + 0 \hat{k} \\ \vec{b} = \hat{i} + 0 \hat{ȷ} + 2 \hat{k} \end{aligned}

$
Also, the second line passes through the point

(1, 3, 0)

and has the directions ratios proportional to

(- 1, 0, 0)

its vector is,

\vec{γ} = \vec{a_{2}} + μ \vec{b_{2}} \to (2)

Here,
$

\begin{aligned} \vec{a_{2}} = \hat{ı} + 3 \hat{j} + 0 \hat{k} \\ \vec{b_{2}} = \hat{ı} + 0 \hat{ȷ} + 0 \hat{k} \end{aligned}

$
Now,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = \hat{ı} + 3 \hat{ȷ} + 0 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 1 & 0 & 2 \\ - 1 & 0 & 0 \end{array} | \Rightarrow 0 \hat{ı} - 2 \hat{ȷ} + 0 \hat{k} \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{0 + 4 + 0} \Rightarrow 2 \\ | (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | = | (\hat{ı} + 3 \hat{ȷ} + 0 \hat{k}) (0 \hat{ı} - 2 \hat{ȷ} + 0 \hat{k}) | \end{aligned}

$
The shortest distance between the lines,
$

\begin{aligned} \vec{γ} & = \vec{a_{1}} + λ \vec{b_{1}} \\ \vec{γ} & = \vec{a_{2}} + μ \vec{b_{2}} \end{aligned}

$
$

\begin{aligned} d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |} \\ d = \frac{| - 6 |}{| 2 |} \\ d = 3 \end{aligned}

Straight Line in Space exercise 27.5 question 5(ii)

\begin{aligned} \frac{x - 0}{1 - 0} = \frac{y - 0}{0 - 0} = \frac{z - 0}{2 - 1} \\ \Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{2} \end{aligned}

$
Solution (ii)
Again applying the points

(1, 3, 0)

and

(0, 3, 0)

The equation of the line passing through the points

(1, 3, 0)

is,
$

\begin{aligned} \frac{x - 1}{0 - 1} = \frac{y - 3}{3 - 3} = \frac{z - 0}{2 - 1} \\ \Rightarrow \frac{x - 1}{- 1} = \frac{y - 3}{0} = \frac{z}{0} \end{aligned}

$
Since the first line passes through the points

(1, 3, 0)

and has the directions proportional to

(1, 0, 2)

its vector is,

\vec{γ} = \vec{a_{1}} + λ \vec{b_{1}} \to (1)

Hence,
$

\begin{aligned} \vec{a} = 0 \hat{ı} + 0 \hat{ȷ} + 0 \hat{k} \\ \vec{b} = \hat{i} + 0 \hat{ȷ} + 2 \hat{k} \end{aligned}

$
Also, the second line passes through the point

(1, 3, 0)

and has the directions ratios proportional to

(- 1, 0, 0)

its vector is,

\vec{γ} = \vec{a_{2}} + μ \vec{b_{2}} \to (2)

Here,
$

\begin{aligned} \vec{a_{2}} = \hat{ı} + 3 \hat{j} + 0 \hat{k} \\ \vec{b_{2}} = \hat{ı} + 0 \hat{ȷ} + 0 \hat{k} \end{aligned}

$
Now,
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) = \hat{ı} + 3 \hat{ȷ} + 0 \hat{k} \\ (\vec{b_{1}} \times \vec{b_{2}}) = | \begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 1 & 0 & 2 \\ - 1 & 0 & 0 \end{array} | \Rightarrow 0 \hat{ı} - 2 \hat{ȷ} + 0 \hat{k} \end{aligned}

$
$

\begin{aligned} | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{0 + 4 + 0} \Rightarrow 2 \\ | (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) | = | (\hat{ı} + 3 \hat{ȷ} + 0 \hat{k}) (0 \hat{ı} - 2 \hat{ȷ} + 0 \hat{k}) | \end{aligned}

$
The shortest distance between the lines,
$

\begin{aligned} \vec{γ} & = \vec{a_{1}} + λ \vec{b_{1}} \\ \vec{γ} & = \vec{a_{2}} + μ \vec{b_{2}} \end{aligned}

$
$

\begin{aligned} d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |} \\ d = \frac{| - 6 |}{| 2 |} \\ d = 3 \end{aligned}

Straight Line in Space exercise 27.5 question 6

Answer: $a_{1} = \hat{ı} + 2 \hat{ȷ} + 4 \hat{k}, b_{1} = 2 \hat{ı} - 3 \hat{ȷ} - 6 \hat{k}$
$a_{2} = 3 \hat{ı} + 3 \hat{ȷ} - 5 \hat{k} b_{2} = 4 \hat{ı} + 6 \hat{ȷ} + 12 \hat{k} and d = \frac{\sqrt{293}}{7} units$

Hint: using expression

| \frac{∣ \vec{a_{2}} - \vec{a_{1})} \times \vec{b_{1}}}{\vec{b}} |

Given: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} and \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{Z + 5}{12}$

Solution:
Let line

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

………… (i)
From above point

(x, y, z)

on line 1 will be

(2 μ + 1, 3 μ + 2, 6 μ - 4)

Let line 2

x - 3 / 4 = \frac{y - 3}{6} = \frac{z + 5}{12} = λ

...............(ii)
From above point

(x, y, z)

on line 2 will be

(4 λ + 3, 6 λ + 3, 12 λ - 5)

Position vector from equation (i) we get
$

\begin{aligned} \vec{8} & = (2 μ + 1) \hat{ı} + (3 μ + 2) \hat{ȷ} + (6 μ - 4) \hat{k} \\ = (\hat{ı} + 2 \hat{ȷ} - 4 \hat{k}) + (2 \hat{ı} + 3 \hat{ȷ} + 6 \hat{k}) \\ \vec{a_{1}} & = \hat{ı} + 2 \hat{ȷ} - 4 \hat{k}, \vec{b_{1}} = 2 \hat{ı} + 3 \hat{ȷ} + 6 \hat{k} \end{aligned}

$
Position vector from evaluation (ii) we get
$

\begin{aligned} \vec{8} & = (4 λ + 3) \hat{ı} + (6 λ + 3) \hat{ȷ} + (12 λ - 5) \hat{k} \\ = (3 \hat{ı} + 3 \hat{ȷ} - 5 \hat{k}) + λ (4 \hat{ı} + 6 \hat{ȷ} + 12 \hat{k} \\ \vec{a_{2}}) & = 3 \hat{ı} + 3 \hat{ȷ} - 5 \hat{k}, \vec{b_{2}} = 4 \hat{ı} + 6 \hat{ȷ} + 12 \hat{k} \end{aligned}

From b_{1} and b_{2} we get \vec{b_{2}} = 2 \vec{b_{1}}

Shortest Distance

= | \frac{∣ \vec{a_{2}} - \vec{a_{1}}) \times \vec{b_{1}}}{\vec{b}} |

(\vec{a_{2}} - \vec{a_{1})} = (3 \hat{i} + 3 \hat{j} - 5 \hat{k}) - (\hat{i} + 2 \hat{ȷ} - 4 \hat{k} = 2 \hat{i} + \hat{j} - \hat{k}

(\vec{a_{2}} - \vec{a_{1})} x \vec{b}

$=\left|

\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 1 \\ 2 & 3 & 6 \end{array}

\right|=9 \hat{i}-14 \hat{\jmath}+4 \widehat{k}$
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1})} \times \vec{b} = \sqrt{19)^{2} + (- 14)^{2} (4)^{2}} = \sqrt{81 + 196 + 16} = \sqrt{293} \\ | \vec{b} | = \sqrt{2^{2} + 3^{2} + 6^{2} = \sqrt{4 + 9 + 36} = 7} \end{aligned}

$
Shortest distance

= \frac{\sqrt{293}}{7}

units

Straight Line in Space exercise 27.5 question 7(i)

Answer: shortest distance

= \frac{3}{\sqrt{2}}

Hints: using expression

D = | \frac{\vec{(a_{2}} - \vec{a_{1})} \vec{(b_{1}} \times \vec{b_{2}})}{(b_{1} \times \overset{―}{b_{2}})} |

Given:

\vec{8} = \hat{ı} + 2 \hat{ȷ} + \hat{k} + λ (\hat{ı} - \hat{ȷ} + \hat{k}) and \vec{8} = 2 \hat{ı} - \hat{ȷ} - \hat{k} + μ (2 \hat{ı} + \hat{ȷ} + 2 \hat{k})

Solution:
$ $\begin{array}{ll} a_{1} = \hat{ı} + 2 \hat{ȷ} + \hat{k}, & b_{1} = \hat{ı} - \hat{ȷ} - \hat{k} \\ a_{2} = 2 \hat{ı} - \hat{ȷ} - \hat{k} & b_{2} = 2 \hat{ı} + \hat{ȷ} + 2 \hat{k} \\ \vec{a_{2}} - \vec{a_{1}} = \hat{ı} - 3 \hat{ȷ} - 2 \hat{k} \end{array}$ $
$\overrightarrow{b_{1}}-\overrightarrow{b_{2}}=\left| $\begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 1 & - 1 & 1 \\ 2 & 1 & 2 \end{array}$ \right|$
$ $\begin{aligned} = \hat{ı} (- 3) - \hat{ȷ} (0) + \hat{k} (1 + 2) \\ = - 3 \hat{ı} + 3 \hat{k} \\ SD = \frac{(2 \hat{i} - \hat{j} - \hat{k}) (- 3 \hat{i} + 3 \hat{k})}{\sqrt{9 + 9}} \end{aligned}$ $
$ $\begin{aligned} = | \frac{6 - 3}{\sqrt{18}} | \\ = \frac{9}{3 \sqrt{2}} \\ = \frac{3}{\sqrt{2}} units \end{aligned}$ $

Straight Line in Space exercise 27.5 question 7(ii)

Answer: $d = 2 \sqrt{29} units$
Hint: using expression $\left|

\begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}

\right|$

\sqrt{{(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2} + {(a_{1} b_{2} + a_{2} b_{1})}^{2}}

Given: $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$
Solution:
It is known that the shortest distance between the two lines,

\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} and \frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}

Comparing the given equation we get,
$

\begin{array}{lll} x_{1} = - 1, & y_{1} = - 1, & z_{1} = - 1 \\ a_{1} = 7, & b_{1} = - 6, & c_{1} = 1 \\ x_{2} = 3, & y_{2} = 5, & z_{2} = 7 \\ a_{2} = 1, & b_{2} = - 2, & c_{2} = 1 \end{array}

$
Then
$

\begin{aligned} | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = | \begin{array}{ccc} 4 & 6 & 8 \\ 7 & - 6 & 1 \\ 1 & - 2 & 1 \end{array} | \\ = 4 (- 6 + 2) - 6 (7 - 1) + 8 (- 14 + 6) = - 116 \end{aligned}

$
$

\begin{aligned} \sqrt{{b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{2})}^{2} + {(a_{1} b_{2} - a_{2} b_{1})}^{2}} \\ = \sqrt{(- 6 + 2)^{2} + (1 + 7)^{2} + (- 14 + 6)^{2}} \\ = \sqrt{16 + 36 + 64} \end{aligned}

$
$

\begin{aligned} = \sqrt{116} \\ = 2 \sqrt{29} \end{aligned}

$
Substituting all the values in equation we get

d = \frac{- 116}{2 \sqrt{29}} = \frac{- 58}{\sqrt{29}} = \frac{- 2 \times 29}{\sqrt{29}} = - 2 \sqrt{29}

Since the distance is always non-negative the distance between the given line is

= 2 \sqrt{29} units

Straight Line in Space exercise 27.5 question 7(iii)

Answer: $d = | \frac{3}{\sqrt{19}} |$
Hint: using

\frac{\vec{(b_{1}} \times \vec{b_{2})} \vec{(a_{2}} - \vec{a_{1}})}{(b_{1} \times \overset{―}{b_{2})}}

Given: $\vec{8} = \hat{ı} + 2 \hat{ȷ} + 3 \hat{k} + λ (\hat{ı} - 3 \hat{ȷ} + 2 \hat{k})$ and

\vec{8} = 4 \hat{ı} + 5 \hat{ȷ} + 6 \hat{k} + μ (2 \hat{ı} + 3 \hat{ȷ} + \hat{k})

Solution:
Shortest distance between the lines with vector equations

\vec{8} = \vec{a_{1}} + \vec{b_{1}} and \vec{8} = \vec{a_{2}} + \vec{b_{2}} is | \frac{\vec{(b_{1}} \times \vec{b_{2})}}{\overset{―}{(b_{1}} \times \overset{―}{a_{2}} - \vec{a_{1}})} |

Now

\vec{8} = \hat{ı} + 2 \hat{ȷ} + 3 \hat{k} + λ (\hat{ı} - 3 \hat{ȷ} + 2 \hat{k})

\vec{8} = (4 \hat{ı} + 5 \hat{ȷ} + 6 \hat{k} + μ (2 \hat{ı} + 3 \hat{ȷ} + \hat{k})

Comparing with

\vec{8} = \vec{a_{1}} + λ \vec{b_{1}}

Comparing with

\vec{8} = \vec{a_{2}} + μ \vec{b_{2}}

\vec{a_{1}} = \hat{i} + 2 \hat{j} + 3 \hat{k}

\vec{a_{2}} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}

and and

\vec{b_{1}} = \hat{i} - 3 \hat{j} + 2 \hat{k}

\vec{b_{2}} = 2 \hat{i} - 3 \hat{j} + 1 \hat{k}

\begin{aligned} \vec{(a_{2}} - \vec{a_{1}}) = (4 \hat{ı} + 5 \hat{ȷ} + 6 \hat{k} - (1 \hat{ı} + 2 \hat{ȷ} + 3 \hat{k}) \\ = (4 - 1) \hat{ı} + (5 - 2) \hat{ȷ} + (6 - 3) \hat{k} \\ = 3 \hat{ı} + 3 \hat{ȷ} + 3 \hat{k} \end{aligned}

$
$\overrightarrow{\left(b_{1}\right.} x \overrightarrow{\left.b_{2}\right)}=\left|

\begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 1 & - 3 & 2 \\ 2 & 3 & 1 \end{array}

\right|$
$

\begin{aligned} = \hat{ı} [(- 3 \times 1) - (3 \times 2)] - \hat{ȷ} [(1 \times 1) - (2 \times 2)] + \hat{k} [(1 \times 3) - (2 \times - 3)] \\ = \hat{ı} (- 9) - \hat{ȷ} (- 3) + \hat{k} (9) \\ = 9 \hat{ı} + 3 \hat{ȷ} + 9 \hat{k} \end{aligned}

$
Magnitude of

\vec{(b_{1}} \times \vec{b_{2})} = \sqrt{(- 9)^{2} + 3^{2} + 9^{2}}

\begin{aligned} = \sqrt{81 + 9 + 81} \\ = \sqrt{171} \end{aligned}

$
$

\begin{aligned} = \sqrt{9 \times 29} \\ = 3 \sqrt{19} \end{aligned}

$
Also
$

\begin{aligned} \vec{(b_{1}} \times \vec{b_{2})} \vec{(a_{2}} - \vec{a_{1}}) = (- 9 (+ 3 \hat{ȷ} + 9 \hat{k}) (3 \hat{ı} + 3 \hat{ȷ} + 3 \hat{k}) \\ = (- 9 \times 3) + (3 \times 3) + (9 + 3) \\ = - 27 + 9 + 27 \\ = 9 \end{aligned}

$
Shortest Distance $

\begin{aligned} = | \frac{9}{3 \sqrt{19}} | \end{aligned}

= | \frac{3}{\sqrt{19}} |

Straight Line in Space exercise 27.5 question 7(iv)

Answer: $d = 9$
Hint: using the expression

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) |}{| (\vec{b_{1}} \times \vec{b_{2}}) |}

Given:

$

\begin{aligned} \vec{8} = 6 \hat{ı} + 2 \hat{ȷ} + 2 \hat{k} + λ (\hat{ı} - 2 \hat{ȷ} + 2 \hat{k}) and \\ \vec{8} = 4 \hat{ı} - \hat{k} + \boldsymbol μ (2 \hat{ı} + 3 \hat{ȷ} + \hat{k}) \end{aligned}

$

Solution:
$

\begin{aligned} \vec{a_{1}} = 6 \hat{ı} + 2 \hat{ȷ} + 2 \hat{k} \\ \vec{a_{2}} = 4 \hat{ı} - \hat{k} \\ \vec{b_{1}} = \hat{ı} - 2 \hat{ȷ} + 2 \hat{k} \end{aligned}

$
$

\begin{aligned} \vec{b_{2}} = {\hat{3}}_{l} - 2 \hat{ȷ} - 2 \hat{k} \\ \vec{a_{2}} - \vec{a_{1}} = - 10 \hat{ı} - 2 \hat{ȷ} + 3 \hat{k} \end{aligned}

$
$\text { and } \overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\left|

\begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 1 & - 2 & 2 \\ 3 & - 2 & - 2 \end{array}

\right| \Rightarrow 8 \hat{\imath}+8 \hat{\jmath}+4 \hat{k}$
$

\begin{aligned} \Rightarrow | \vec{b_{1}} \times \overset{―}{b_{2}} | \Rightarrow \sqrt{8^{2} + 8^{2} + 4^{2}} \Rightarrow \sqrt{64 + 64 + 16} \\ = \sqrt{144} \Rightarrow 12 \end{aligned}

$
and
$

\begin{aligned} (\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}}) = - 10 \hat{ı} - 2 \hat{ȷ} + 3 \hat{k} (8 \hat{ı} + 8 \hat{ȷ} + 4 \hat{k}) \\ = - 80 - 16 - 12 \\ = - 108 \end{aligned}

$
The shortest distance between the lines
$

\begin{aligned} \vec{8} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{8} = \vec{a_{2}} + μ b_{2} \\ d = | \frac{- 108}{12} | \\ = 9 \end{aligned}

Straight Line in Space exercise 27.5 question 8

Answer: $D = \frac{1}{7}$
Hint: using formula distance between

∥

line

| \frac{(\vec{a_{2}} - \vec{a_{1}}) \vec{b}}{\vec{b}} |

Given: $\vec{8} = \hat{ı} + 2 \hat{ȷ} - 4 \hat{k} + + λ (2 \hat{ı} - 3 \hat{ȷ} + 6 \hat{k}) and$
$\vec{8} = 3 \hat{ı} + 3 \hat{ȷ} - 5 \hat{k} + μ (2 \hat{ı} + 3 \hat{ȷ} + 6 \hat{k})$
Solution:
By the given formula,
$

\begin{aligned} = | \frac{(2 \hat{ı} + \hat{ȷ} - \hat{k}) (2 \hat{i} + 3 \hat{j} + 6 \hat{k})}{\sqrt{4 + 9 + 36}} | \\ = \frac{4 + 3 - 6}{\sqrt{49}} \end{aligned}

$
$

\begin{aligned} = \frac{4 + 3 - 6}{7} \\ = 1 / 7 \end{aligned}

Straight Line in Space exercise 27.5 question 9

Answer: $d = \frac{\sqrt{580}}{7} units$
Hint: find l₂ by the given points.
Given: $\vec{8} = (- 2 \hat{ı} + 3 \hat{ȷ}) + λ (2 \hat{ı} - 3 \hat{ȷ} + 6 \hat{k}), point (2, 3, 2)$
Solution:
Evaluation of line passing through $(2, 3, 2)$ and parallel to $\vec{2}$ is given by
$L_{2} = (2 \hat{ı} + 3 \hat{ȷ} + 2 \hat{k}) + μ (2 \hat{ı} - 3 \hat{ȷ} + 6 \hat{k})$
Now,
$(\vec{a_{2}} - a_{1}) = (μ \hat{ı} + 2 \hat{k})$
and $\vec{5} = (2 \hat{ı} - 3 \hat{ȷ} + 6 \hat{k})$
Now,
$\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) x \overrightarrow{b_{2}}=\left| $\begin{array}{ccc} \hat{ı} & \hat{ȷ} & \hat{k} \\ 4 & 0 & 2 \\ 2 & - 3 & 6 \end{array}$ \right| \quad 6 \hat{\imath}-20 \hat{\jmath}+12 \hat{k}$
Hence the revised distance is
$\frac{(\vec{a_{2}} - \vec{a_{1}}) \times \vec{b}}{(\vec{b})} = \frac{\sqrt{36 + 400 + 144}}{\sqrt{4 + 9 + 36}}$
$= \frac{\sqrt{580}}{7} units$

The RD Sharma Class 12 Solution of Straight Line in Space exercise 27.5 is used by thousands of students and teachers for the practical knowledge of maths. The RD Sharma class 12th exercise 27.5 consists of 27 questions covering concepts like shortest distance between pairs of lines in vector and cartesian form of the equation, vector equation of a line passing through the point, and parallel to the bar.

The benefits of practicing from the RD Sharma class 12th exercise 27.5 are mentioned below: -

RD Sharma class 12th exercise 27.5 Is prepared by a team of subject experts who have years of experience with the CBSE exam paper pattern. This material will help students get a better insight into the subject and score good marks in exams.
Students can utilize RD Sharma class 12th exercise 27.5 material as a guide for their RD Sharma textbook. As it contains questions and answers in the same place, it is convenient for students for revision purposes.
Career360 provides RD Sharma class 12 solutions chapter 27 exercise 27.5 for free on their website. This means that students can access them to any device with an internet connection. This material is made to help students get a head start on the subject and reduce the number of doubts.

RD Sharma class 12th exercise 27.5 material is ready by specialists who have long stretches of involvement and understanding on exam designs. These solutions are intended to furnish students with the best wellspring of readiness and get them to abound together material for every one of their chapters.

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RD Sharma Class 12 Exercise 27.5 Straight line in space Solutions Maths - Download PDF Free Online

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Straight Line in Space Excercise: 27.5

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