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: $\text { } \overrightarrow{\mathrm{\gamma}}=(3 \hat{\imath}+8 \hat{\jmath}+3 \hat{k})+\lambda(3 \hat{\imath}-\hat{\jmath}+\hat{k}) \text { and }(-3 \hat{\imath}-\hat{\jmath}+\hat{k})+\mu(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})$

Solution: From the given pair of equation
$\begin{aligned} &\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1} \Rightarrow 1 \\\\ &\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{-4} \Rightarrow 2 \end{aligned}$
$d=\frac{\left|\begin{array}{ccc} 6 & 15 & -3 \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{array}\right|}{\sqrt{(6-3)^{2}+(-4-2)^{2}+(12+3)^{2}}}$
$\begin{aligned} &=\left|\frac{6(-4-2)-15(12+3)-3(6-3)}{\sqrt{9+36+225}}\right| \\\\ &=\frac{-36-225-9}{\sqrt{270}} \end{aligned}$
$\begin{aligned} &=\frac{270}{\sqrt{270}} \\\\ &=\sqrt{270} \\\\ &=3 \sqrt{30} \Rightarrow \mathrm{ANS} \end{aligned}$

Straight Line in Space exercise 27.5 question 1(ii)

Answer: $D=\frac{64}{\sqrt{62}} \text { units }$
Hint: Using formula $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overline{b_{2}}\right)\right|}$

Given: $\overrightarrow{\mathrm{\gamma}}=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}+7 \hat{k}) \text { and } \overrightarrow{\mathrm{\gamma}}=(-\hat{\imath}-\hat{\jmath}-\hat{k})+\mu(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})$

Solution:
$\begin{aligned} &\overrightarrow{a_{1}}=3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k} \\\\ &\overrightarrow{a_{2}}=-\hat{\imath}-\hat{\jmath}-\hat{k} \\\\ &\overrightarrow{b_{1}}=\hat{\imath}-2 \hat{\jmath}+7 \hat{k} \\\\ &\overrightarrow{b_{2}}=7 \hat{\imath}-6 \hat{\jmath}+\hat{k} \end{aligned}$
The shortest distance between the lines is
$\begin{gathered} d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overline{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|} \\\\ \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(-\hat{\imath}-\hat{\jmath}-\hat{k})-(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k}) \\\\ =-4 \hat{\imath}-6 \hat{\jmath}-8 \hat{k} \end{gathered}$
$\begin{aligned} \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -2 & 7 \\ 7 & -6 & 1 \end{array}\right| \\\\ \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=(-2+42)+[-(1-49) \hat{j}]+(-6+14 \hat{k}) \\\\ &=40 \hat{\imath}+48 \hat{\jmath}+8 \hat{k} \end{aligned}$
$\begin{aligned} \left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right| &=\sqrt{(40)^{2}+(48)^{2}+(8)^{2}} \\\\ &=8 \sqrt{62} \end{aligned}$
$\begin{aligned} \left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right| &=|(-4 \hat{\imath}-6 \hat{\jmath}-8 \hat{k})(40 \hat{\imath}+48 \hat{\jmath}+8 \hat{k})| \Rightarrow \\\\ \left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right| &=|-512| \\\\ &=512 \end{aligned}$
Putting the values in the above mentioned formula,
$\begin{aligned} &d=\frac{512}{8 \sqrt{62}} \\\\ &=\frac{64}{\sqrt{62}} \\\\ &=\frac{64}{\sqrt{62}} \text { units } \end{aligned}$

Straight Line in Space exercise 27.5 question 1(iii)

Answer: $d=\frac{1}{\sqrt{6}} \text { units }$
Hint: Using formula $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$

Given: $\vec{\gamma}=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}) \text { and } \overrightarrow{\mathrm{\gamma}}=(2 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})+\mu(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})$
Solution:
By the given pairs of equation,
$\begin{aligned} &\overrightarrow{a_{1}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \\\\ &\overrightarrow{a_{2}}=2 \hat{\imath}+4 \hat{j}+5 \hat{k} \\\\ &\overrightarrow{b_{1}}=2 \hat{\imath}+3 \hat{\jmath}+5 \hat{k} \\\\ &\overrightarrow{b_{2}}=3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k} \end{aligned}$
The shortest distance between the lines is,
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(2 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=\hat{\imath}+2 \hat{\jmath}+2 \hat{k} \end{aligned}$
$\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right| \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=-\hat{\imath}+2 \hat{\jmath}-\hat{k} \end{aligned}$
$\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-1)^{2}+(2)^{2}+(-1)^{2}} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{6} \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(\hat{\imath}+2 \hat{\jmath}+2 \hat{k})(-\hat{\imath}+22 \hat{\jmath}-\hat{k})| \Rightarrow 1 \end{aligned}$
Putting three values in the expression,
$\begin{aligned} &d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|} \\\\ &d=\frac{1}{\sqrt{6}} \text { units } \end{aligned}$

Straight Line in Space exercise 27.5 question 1(iv)

Answer: $d=\frac{8}{\sqrt{29}} \text { units }$
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: $\overrightarrow{\mathrm{\gamma}}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-t) \hat{k} \text { and } \overrightarrow{\mathrm{\gamma}}=(s+) \hat{\imath}+(2 s-1) \hat{\jmath}-(2 s+1) \hat{k}$
Solution:
$\begin{array}{ll} \overrightarrow{\mathrm{\gamma}}=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+t(-\hat{\imath}+\hat{\jmath}-2 \hat{k}) & {[\overrightarrow{\mathrm{\gamma}}=\vec{a}+\lambda \vec{b}]} \\\\ \overrightarrow{\mathrm{\gamma}}=(\hat{\imath}-\hat{\jmath}-\hat{k})+s(\hat{\imath}+2 \hat{\jmath}-2 \hat{k}) & {[\overrightarrow{\mathrm{\gamma}}=\vec{c}+\lambda \vec{d}]} \end{array}$
$\vec{b} \times \vec{d}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{array}\right|$
$=\hat{\imath}[1(-2)-(-2)(2)]-\hat{\jmath}[(-1)(-2)-(1)(-2)]+\hat{k}[(-1)(2)-(1)(1)]$
$\begin{aligned} &\vec{b} \times \vec{d}=2 \hat{\imath}-4 \hat{\jmath}-3 \hat{k} \\\\ &|\vec{b} \times \vec{d}|=\sqrt{(2)^{2}+(-4)^{2}+(-3)^{2}} \Rightarrow \sqrt{9+16+4} \Rightarrow \sqrt{29} \end{aligned}$
$\begin{aligned} (\vec{c}-\vec{a}) &=(\hat{\jmath}-4 \hat{k}) \\\\ d &=\frac{|(\hat{\jmath}-4 \hat{k})(2 \hat{\imath}-4 \hat{\jmath}-3 \hat{k})|}{|\sqrt{29}|} \Rightarrow \frac{|0-4+12|}{|\sqrt{29}|} \Rightarrow \frac{8}{\sqrt{29}} \end{aligned}$

Straight Line in Space exercise 27.5 question 1(v)

Answer: $d=\frac{5}{2} \sqrt{2} \text { units }$
Hint: Convert the given equation in the normal form $\overrightarrow{\mathrm{\gamma }}=\vec{a}+\lambda \vec{b}$
$\begin{aligned} \text { Let } \overrightarrow{\mathrm{\gamma}} &=a_{1}+\lambda b_{1} \\\\ \overrightarrow{\mathrm{\gamma}} &=a_{2}+\mu b_{2} \end{aligned}$
Given: $\vec{\gamma}=(\lambda-1) \hat{\imath}+(\lambda+1) \hat{\jmath}-(1+\lambda) \hat{k} \text { and } \overrightarrow{\mathrm{\gamma}}=(1-\mu) \hat{\imath}+(2 \mu-1) \hat{\jmath}+(\mu+2) \hat{k}$
Solution:
$\begin{aligned} &\vec{\gamma}=(-\hat{\imath}+\hat{\jmath}-\hat{k})+\lambda(\hat{\imath}+\hat{\jmath}-\hat{k}) \text { and } \\\\ &\vec{\gamma}=(\hat{\imath}-\hat{\jmath}+2 \hat{k})+\mu(-\hat{\imath}+2 \hat{\jmath}+\hat{k}) \\\\ &D=\frac{\left|\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)\right|}{\left|\left(b_{1} \times b_{2}\right)\right|} \end{aligned}$
Here
$\begin{array}{ll} \overrightarrow{a_{1}}=-\hat{\imath}+\hat{\jmath}-\hat{k} & \overrightarrow{b_{1}}=\hat{\imath}+\hat{\jmath}-\hat{k} \\\\ \overrightarrow{a_{2}}=\hat{\imath}-\hat{\jmath}+2 \hat{k} & \overrightarrow{b_{2}}=-\hat{\imath} \\\\ \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=2 \hat{i}-2 \hat{\jmath}+3 \hat{k} & \end{array}$
$\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & -1 \\ -1 & 2 & 1 \end{array}\right|$
$\begin{aligned} \overrightarrow{b_{1}} \times \overrightarrow{b_{2}} &=\hat{\imath}(1+2)-\hat{\jmath}(-1-1)+\hat{k}(2+1) \\\\ &=3 \hat{\imath}+3 \hat{k} \end{aligned}$
$\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(3)^{2}+(3)^{2}} \Rightarrow 3 \sqrt{2} \\\\ &\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)=(2 \hat{\imath}-2 \hat{\jmath}+3 \hat{k}) \cdot(3 \hat{\imath}+3 \hat{k}) \Rightarrow 6+9 \Rightarrow 15 \\\\ &D=\frac{15}{3 \sqrt{2}} \Rightarrow \frac{15}{\sqrt{2}} \Rightarrow \frac{5}{2 \sqrt{2}} \end{aligned}$

Straight Line in Space exercise 27.5 question 1(vi)

Answer: $d=\frac{3}{\sqrt{2}} \text { units }$
Hint: Using the expression $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$

Given: $\overrightarrow{\mathrm{\gamma}}=(2 \hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(2 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}) \text { and } \overrightarrow{\mathrm{\gamma}}=(\hat{\imath}+2 \hat{\boldsymbol{j}}+\hat{k})+\mu(\hat{\imath}-\hat{\jmath}+\hat{k})$
Solution:
$\begin{array}{ll} \overrightarrow{a_{1}}=2 \hat{\imath}-\hat{\jmath}-\hat{k} & \overrightarrow{b_{1}}=2 \hat{\imath}-5 \hat{\jmath}+2 \hat{k} \\\\ \overrightarrow{a_{2}}=\hat{\imath}+2 \hat{\jmath}+\hat{k} & \overrightarrow{b_{2}}=\hat{\imath}-\hat{\jmath}+\hat{k} \end{array}$
The shortest distance between the lines is,
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(\hat{\imath}+2 \hat{\jmath}+\hat{k})-(2 \hat{\imath}-\hat{\jmath}-\hat{k}) \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=-\hat{\imath}+3 \hat{\jmath}+2 \hat{k} \end{aligned}$
$\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -5 & 2 \\ 1 & -1 & 1 \end{array}\right|$
$\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-5+2) \hat{\imath}-(2-2) \hat{\jmath}+(-2+5) \hat{k} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-3)^{2}+(0)^{2}+(3)^{2}} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=3 \sqrt{2} \end{aligned}$
$\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})(-3 \hat{\imath}+0 \hat{\jmath}+3 \hat{k})| \Rightarrow 9$
Putting three values in the above mentioned equation,
$\begin{aligned} &d=\frac{9}{3 \sqrt{2}} \\\\ &d=\frac{3}{\sqrt{2}} \text { units } \end{aligned}$

Straight Line in Space exercise 27.5 question 1(vii)

Answer: $d=\frac{10}{\sqrt{59}} \text { units }$
Hint: Using the expression $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$
Given: $\vec{\gamma}=(\hat{\imath}+\hat{\jmath})+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \text { and } \vec{\gamma}=(2 \hat{\imath}+\hat{\jmath}-\hat{k})+\mu(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k})$
Solution: We know that,
$\begin{array}{ll} \overrightarrow{a_{1}}=\hat{\imath}+\hat{\jmath} & \overrightarrow{b_{1}}=2 \hat{\imath}-\hat{\jmath}+\hat{k} \\\\ \overrightarrow{a_{2}}=2 \hat{\imath}+\hat{\jmath}-\hat{k} & \overrightarrow{b_{2}}=3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k} \end{array}$
It can be written as,
$\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{array}\right| \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-2+5) \hat{\imath}-(4-3) \hat{\jmath}+(-10+3) \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=3 \hat{\imath}-\hat{\jmath}-7 \hat{k} \end{aligned}$
$\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(3)^{2}+(-1)^{2}+(-7)^{2}} \Rightarrow \sqrt{9+1+49} \Rightarrow \sqrt{59} \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(2-1) \hat{\imath}-(1-1) \hat{\jmath}+(-1-0) \hat{k}=\hat{\imath}+0 \hat{\jmath}-\hat{k} \end{aligned}$
We can write as,
$\begin{aligned} \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) &=(3 \hat{\imath}-\hat{\jmath}-7 \hat{k}) \cdot(\hat{\imath}+0 \hat{\jmath}-\hat{k}) \\\\ &=(3 \times 1)+[(-1) \times 0]+([(-7) \times(-1)\\\\ &=3+0+7 \Rightarrow 10 \end{aligned}$
The shortest distance between the given lines is,
$d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|} \Rightarrow \frac{10}{\sqrt{59}} \text { units }$

Straight Line in Space exercise 27.5 question 1(viii)

Answer: $d=14 \text { units }$
Hint: Using the expression to find $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$

Given: $\overrightarrow{\mathrm{\gamma }}=(8+3 \lambda) \hat{\imath}-(9+16 \lambda) \hat{\jmath}+(10+7 \lambda) \hat{k} \text { and } \overrightarrow{\mathrm{\gamma }}=(15 \hat{\imath}+29 \hat{\jmath}+5 \hat{k})+$ $\mu(3 \hat{\imath}+8 \hat{\jmath}-2 \hat{k})$

Solution: We know that,
$\begin{array}{ll} \overrightarrow{a_{1}}=8 \hat{\imath}-9 \hat{\jmath}+10 \hat{k} & \overrightarrow{b_{1}}=3 \hat{\imath}-16 \hat{\jmath}+7 \hat{k} \\\\ \overrightarrow{a_{2}}=15 \hat{\imath}+29 \hat{\jmath}+5 \hat{k} & \overrightarrow{b_{2}}=3 \hat{\imath}+8 \hat{\jmath}-5 \hat{k} \end{array}$
The shortest distance between the given lines is,
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(15 \hat{\imath}+29 \hat{\jmath}+5 \hat{k})-(8 \hat{\imath}-9 \hat{\jmath}-10 \hat{k}) \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=7 \hat{\imath}+38 \hat{\jmath}-5 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & -16 & 7 \\ 3 & -8 & -5 \end{array}\right| \end{aligned}$
$\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(80-56) \hat{\imath}-(-15-21) \hat{\jmath}+(24+48) \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=24 \hat{\imath}+36 \hat{\jmath}+72 \hat{k} \Rightarrow 4(6 \hat{\imath}+8 \hat{\jmath}+18 \hat{k}) \end{aligned}$
$\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=4 \sqrt{(6)^{2}+(9)^{2}+(18)^{2}} \Rightarrow 4 \sqrt{441} \Rightarrow 4 \times 21 \Rightarrow 84 \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|4(7 \hat{\imath}+38 \hat{\jmath}-5 \hat{k})(6 \hat{\imath}+9 \hat{\jmath}+18 \hat{k})| \Rightarrow 4(42+342-90) \Rightarrow 1176 \end{aligned}$
Putting three values in the above mentioned equation,
$\begin{aligned} &d=\frac{1176}{84} \\\\ &d=14 \text { units } \end{aligned}$

Straight Line in Space exercise 27.5 question 2(ii)

Answer: $d=\frac{1}{\sqrt{6}} \text { units }$
Hint: Using the expression to find $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$
Given: $: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \rightarrow(1) \text { and } \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{4} \rightarrow(2)$
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 $\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$
Here,
$\begin{aligned} &\overrightarrow{a_{1}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \\\\ &\overrightarrow{b_{1}}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \end{aligned}$
Also line (2) passes through the point $(2,3,5)$ and has direction ratios proportional to $(3,4,5)$ its vector equation is $\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$
Here,
$\begin{aligned} &\overrightarrow{a_{2}}=2 \hat{\imath}+3 \hat{\jmath}+5 \hat{k} \\\\ &\overrightarrow{b_{2}}=3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k} \end{aligned}$
Now,
$\begin{aligned} \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) &=\hat{\imath}+\hat{j}+2 \hat{k} \\\\ \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right| \\\\ &=-\hat{\imath}+2 \hat{\jmath}-\hat{k} \end{aligned}$
$\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-1)^{2}+(2)^{2}+(-1)^{2}} \Rightarrow \sqrt{6} \text { and } \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(\hat{\imath}+\hat{\jmath}+2 \hat{k})(-\hat{\imath}+2 \hat{\jmath}-\hat{k}) \Rightarrow-1+2-2 \Rightarrow-1 \end{aligned}$
The shortest distance between the given lines is,
$d=\left|\frac{-1}{\sqrt{6}}\right| \Rightarrow \frac{1}{\sqrt{6}} \text { units }$

Straight Line in Space exercise 27.5 question 2(ii)

Answer: $d=\frac{3}{\sqrt{59}} \text { units }$
Hint: Consider the given equation as (1) and (2) and using the expression $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$
Given: $: \frac{x-1}{2}=\frac{y+1}{3}=z \rightarrow(1) \text { and } \frac{x+1}{3}=\frac{y-2}{1} ; z=2 \rightarrow(2)$
Solution:
Since line (1) passes through the point $(1,-1,0)$ and has direction ratios proportional to $(2,3,1)$ its vector equation is $\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$
Here,
$\begin{aligned} &\overrightarrow{a_{1}}=\hat{\imath}-\hat{\jmath}+0 \hat{k} \\\\ &\overrightarrow{b_{1}}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k} \end{aligned}$
Also line (2) passes through the point $(-1,2,2)$ and has direction ratios proportional to $(3,1,0)$ its vector equation is $\vec{\gamma}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$
Here,
$\begin{aligned} &\overrightarrow{a_{2}}=-\hat{\imath}+2 \hat{\jmath}+2 \hat{k} \\\\ &\overrightarrow{b_{2}}=3 \hat{\imath}+\hat{\jmath}+0 \hat{k} \end{aligned}$
Now,
$\begin{aligned} \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) &=-2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \\\\ \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 1 \\ 3 & 1 & 0 \end{array}\right| \\\\ &=-\hat{\imath}+3 \hat{\jmath}-7 \hat{k} \end{aligned}$
$\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-1)^{2}+(3)^{2}+(-7)^{2}} \Rightarrow \sqrt{1+9+49} \Rightarrow \sqrt{59} \text { and } \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k})(-\hat{\imath}+3 \hat{\jmath}-7 \hat{k}) \Rightarrow 2+9-14 \Rightarrow-3 \end{aligned}$
The shortest distance between the lines $\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \text { and } \overrightarrow{\mathrm{V}}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$ is given by
$d=\left|\frac{-3}{\sqrt{59}}\right| \Rightarrow \frac{3}{\sqrt{59}} \text { units }$

Straight Line in Space exercise 27.5 question 2(iii)

Answer: $d=\frac{8}{\sqrt{29}} \text { units }$
Hint: Consider the given equation as (1) and (2)
Given: $\frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2} \rightarrow(1) \text { and } \frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2} \rightarrow(2)$
Solution: Since line (1) passes through the point $(1,-2,3)$ and has direction ratios proportional to $(-1,1,-2)$ its vector equation is $\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$
Here,
$\begin{aligned} &\overrightarrow{a_{1}}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k} \\\\ &\overrightarrow{b_{1}}=-\hat{\imath}+\hat{\jmath}-2 \hat{k} \end{aligned}$
Also line (2) passes through the point $(1,-1,1)$ and has direction ratios proportional to $(1,2,-2)$ its vector equation is $\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$
Here,
$\begin{aligned} &\overrightarrow{a_{2}}=\hat{\imath}-\hat{\jmath}-\hat{k} \\\\ &\overrightarrow{b_{2}}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k} \end{aligned}$
Now,
$\begin{aligned} \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) &=\hat{\jmath}-2 \hat{k} \\\\ \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{array}\right| \\\\ &=2 \hat{\imath}-u \hat{\jmath}-3 \hat{k} \end{aligned}$
$\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(2)^{2}+(-4)^{2}+(-3)^{2}} \Rightarrow \sqrt{4+16+9} \Rightarrow \sqrt{29} \text { and }$
The shortest distance between the lines
$\begin{aligned} &d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|} \\\\ &d=\left|\frac{8}{\sqrt{29}}\right| \Rightarrow \frac{8}{\sqrt{29}} \text { units } \end{aligned}$

Straight Line in Space exercise 27.5 question 2(iv)

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

Now, let’s take a point on first line as $A(\lambda+3,2 \lambda+5, \lambda+7)$ and let $B(7 k-1,-6 k-1, k-1)$ be point on the second line.
The direction ratio of the line;
$(7 k-\lambda-4) \times 1+(-6 k+2 \lambda-6) \times(-2)+(k-\lambda-8) \times 1=0 \text { Line (1) and }$
$(7 k-\lambda-4) \times 7+(-6 k+2 \lambda-6) \times(-6)+(k-\lambda-8) \times 1=0 \text { Line(2) }$
Solving equation (1) and (2) we get
$\begin{aligned} &\lambda=0 \text { and } k=0 \\\\ &A=(3,5,7) \text { and } B=(-1,-1,-1) \\\\ &A B=\sqrt{(3+1)^{2}+(5+1)^{2}+(7+1)^{2}} \Rightarrow \sqrt{16+32+64} \Rightarrow \sqrt{116} \Rightarrow 2 \sqrt{29} \text { units } \end{aligned}$

Straight Line in Space exercise 27.5 question 3(i)

Answer: Given Lines are not interesting
Hint: Using the equation $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$
Given: $\vec{\gamma}=(\hat{\imath}-\hat{\jmath})+\lambda(2 \hat{\imath}+\hat{k}) \text { and } \overrightarrow{\mathrm{\gamma}}=(2 \hat{\imath}-\hat{\jmath})+\mu(\hat{\imath}+\hat{\jmath}+\hat{k})$
Solution:
We know that,
$\begin{array}{ll} \overrightarrow{a_{1}}=\hat{\imath}-\hat{\jmath}+0 \hat{k} & \overrightarrow{b_{1}}=2 \hat{\imath}+0 \hat{\jmath}+\hat{k} \\\\ \overrightarrow{a_{2}}=2 \hat{\imath}-\hat{\jmath} & \overrightarrow{b_{2}}=\hat{\imath}+\hat{\jmath}-\hat{k} \end{array}$
The shortest distance between the lines is,
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(2 \hat{\imath}-\hat{\jmath})-(\hat{\imath}-\hat{\jmath}+0 \hat{k}) \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=\hat{\imath}+0 \hat{\jmath}+0 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 0 & 1 \\\\ 1 & 1 & -1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(0-1) \hat{\imath}-(-2-1) \hat{\jmath}+(2-0) \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=-\hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-1)^{2}+(3)^{2}+(2)^{2}} \Rightarrow \sqrt{14} \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(\hat{\imath}+0 \hat{\jmath}+0 \hat{k})(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})| \Rightarrow 1 \end{aligned}$
Putting three values in the above mentioned equation,
$\begin{aligned} &d=\frac{1}{\sqrt{14}} \\\\ &d=\frac{1}{\sqrt{14}} \text { units } \end{aligned}$
∴ 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 $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$
Given: $\overrightarrow{\mathrm{\gamma }}=(\hat{\imath}+\hat{\jmath}-\hat{k})+\lambda(3 \hat{\imath}-\hat{\jmath}) \text { and } \overrightarrow{\mathrm{\gamma}}=(4 \hat{\imath}-\hat{k})+\mu(2 \hat{\imath}+3 \hat{k})$
Solution:
We know that,
$\begin{array}{ll} \overrightarrow{a_{1}}=\hat{\imath}+\hat{\jmath}-\hat{k} & \overrightarrow{b_{1}}=3 \hat{\imath}-1 \hat{\jmath}+0 \hat{k} \\\\ \overrightarrow{a_{2}}=4 \hat{\imath}+0 \hat{\jmath}-\hat{k} & \overrightarrow{b_{2}}=2 \hat{\imath}+0 \hat{\jmath}+3 \hat{k} \end{array}$
The shortest distance between the lines is,
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(4 \hat{\imath}+0 \hat{\jmath}-\hat{k})-(\hat{\imath}+\hat{\jmath}-\hat{k}) \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=3 \hat{\imath}-\hat{\jmath}+0 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & -1 & 0 \\ 2 & 0 & 3 \end{array}\right| \end{aligned}$
$\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(3-0) \hat{\imath}-(9-0) \hat{\jmath}+(0+2) \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=-3 \hat{\imath}-9 \hat{j}+2 \hat{k} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-3)^{2}+(-9)^{2}+(2)^{2}} \Rightarrow \sqrt{94} \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(3 \hat{\imath}-1 \hat{\jmath}+0 \hat{k})(-3 \hat{\imath}-9 \hat{\jmath}+2 \hat{k})| \Rightarrow 0 \end{aligned}$
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: $: \frac{x-1}{2}=\frac{y+1}{3}=z \rightarrow(1) \text { and } \frac{x+1}{5}=\frac{y-2}{1} ; z=2 \rightarrow(2)$
Solution: Since line (1) passes through the point $(1,-1,0)$ and has direction ratios proportional to $(2,3,1)$ its vector equation is $\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$
Here
$\begin{aligned} &\overrightarrow{a_{1}}=\hat{\imath}-1 \hat{\jmath}+0 \hat{k} \\\\ &\overrightarrow{b_{1}}=2 \hat{i}+3 \hat{j}+\hat{k} \end{aligned}$
Also line (2) passes through the point $(-1,2,2)$ and has direction ratios proportional to $(5,1,0)$ its vector equation is $\overrightarrow{\mathrm{\gamma }}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$
Here,
$\begin{aligned} &\overrightarrow{a_{2}}=-\hat{\imath}+2 \hat{\jmath}+2 \hat{k} \\\\ &\overrightarrow{b_{2}}=5 \hat{i}+1 \hat{j}+0 \hat{k} \end{aligned}$
Now,
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=-2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 0 \end{array}\right| \Rightarrow-\hat{\imath}+5 \hat{\jmath}-13 \hat{k} \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k}) \cdot(-\hat{\imath}+5 \hat{\jmath}-13 \hat{k}) \Rightarrow 9+15-26 \Rightarrow 9 \end{aligned}$
We observe, $\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \neq 0$
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: $\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5} \rightarrow(1) \text { and } \frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3} \rightarrow(2)$
Solution: Since line (1) passes through the point $(5,7,-3)$ and has direction ratios proportional to $(4,-5,-5)$ its vector equation is $\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$
Here,
$\begin{aligned} &\overrightarrow{a_{1}}=5 \hat{\imath}+7 \hat{\jmath}-3 \hat{k} \\\\ &\overrightarrow{b_{1}}=4 \hat{\imath}-5 \hat{\jmath}-5 \hat{k} \end{aligned}$
Also line (2) passes through the point $(8,7,5)$ and has direction ratios proportional to $(7,1,3)$ its vector equation is $\overrightarrow{\mathrm{\gamma }}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$
Here,
$\begin{aligned} &\overrightarrow{a_{2}}=8 \hat{\imath}+7 \hat{\jmath}+5 \hat{k} \\\\ &\overrightarrow{b_{2}}=7 \hat{i}+1 \hat{\jmath}+3 \hat{k} \end{aligned}$
Now,
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=3 \hat{\imath}+8 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4 & -5 & -5 \\ 7 & 1 & 3 \end{array}\right| \Rightarrow-10 \hat{\imath}-47 \hat{\jmath}+39 \hat{k} \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(3 \hat{\imath}+8 \hat{k}) \cdot(-10 \hat{\imath}-47 \hat{\jmath}+39 \hat{k}) \Rightarrow-30+312 \Rightarrow 282 \end{aligned}$
We observe, $\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \neq 0$
Thus, the given lines are not intersecting.

Straight Line in Space exercise 27.5 question 4(i)

Answer: $d=\sqrt{26} \text { units }$
Hint: Let in $L2$ these is λ draw $\mid \mid$ Line

Given: $\overrightarrow{\mathrm{\gamma }}=(1 \hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(1 \hat{\imath}-1 \hat{\jmath}+\hat{k}) \text { and } \overrightarrow{\mathrm{\gamma }}=(2 \hat{\imath}-\hat{\jmath}-\hat{k})+\mu(-\hat{\imath}+\hat{\jmath}-\hat{k})$
Solution:
$\begin{aligned} &L_{1}(1,2,3), \quad \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-3}{1} \\\\ &L_{2}(2,-1,-1), \frac{x-2}{1}=\frac{y+1}{-1}=\frac{z+1}{1}=\lambda \end{aligned}$
Line (2) becomes $(\lambda+2,-\lambda-1, \lambda-1)$
$\begin{aligned} &=(\lambda+2-1) \hat{\imath}+(-\lambda-1-2) \hat{\jmath}+(\lambda-1-3) \hat{k} \\\\ &=(\lambda+1) \hat{\imath}+(-\lambda-3) \hat{\jmath}+(\lambda-4) \hat{k} \end{aligned}$
$\begin{aligned} &=(\lambda+1)(1)+(-\lambda-3)(-1)+(\lambda-4)(1) \\\\ &=\lambda+1+\lambda+3+\lambda-4 \\\\ &=3 \lambda=0 \end{aligned}$
$\therefore \lambda=0$
The shortest distance between the $\mid \mid$ Line is $2,-1,-1 \; and \; (1,2,3)$

Straight Line in Space exercise 27.5 question 4(ii)

Answer: $d=0$
Hint:Using the equation $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$ Let the given lines L₁ and L₂
Given: $\overrightarrow{\mathrm{\gamma}}=(\hat{\imath}+\hat{\jmath})+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \text { and } \overrightarrow{\mathrm{\gamma }}=(2 \hat{\imath}+\hat{\jmath}-\hat{k})+\mu(4 \hat{\imath}-2 \hat{\jmath}+2 \hat{k})$
Solution:
$\begin{aligned} &L_{1}=a_{1}+\lambda b_{1} \text { and } L_{2}=a_{2}+\mu b_{2} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 1 \\ 4 & -2 & 2 \end{array}\right| \Rightarrow \hat{\imath}(0)-\hat{\jmath}(0)+\hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=0 \end{aligned}$
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: $(0,0,0) \text { and }(1,0,2)$
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,
$\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \rightarrow(1)$
Hence,
$\begin{aligned} &\vec{a}=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k} \\\\ &\vec{b}=\hat{i}+0 \hat{\jmath}+2 \hat{k} \end{aligned}$
Also, the second line passes through the point $(1,3,0)$ and has the directions ratios proportional to $(-1,0,0)$ its vector is,
$\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} \rightarrow \text { (2) }$
Here,
$\begin{aligned} &\overrightarrow{a_{2}}=\hat{\imath}+3 \hat{j}+0 \hat{k} \\\\ &\overrightarrow{b_{2}}=\hat{\imath}+0 \hat{\jmath}+0 \hat{k} \end{aligned}$
Now,
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=\hat{\imath}+3 \hat{\jmath}+0 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 0 & 2 \\ -1 & 0 & 0 \end{array}\right| \Rightarrow 0 \hat{\imath}-2 \hat{\jmath}+0 \hat{k} \end{aligned}$
$\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{0+4+0} \Rightarrow 2 \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(\hat{\imath}+3 \hat{\jmath}+0 \hat{k})(0 \hat{\imath}-2 \hat{\jmath}+0 \hat{k})| \end{aligned}$
The shortest distance between the lines,
$\begin{aligned} \overrightarrow{\mathrm{\gamma }} &=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \\\\ \overrightarrow{\mathrm{\gamma }} &=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} \end{aligned}$
$\begin{aligned} &d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|} \\\\ &d=\frac{|-6|}{|2|} \\\\ &d=3 \end{aligned}$

Straight Line in Space exercise 27.5 question 5(ii)

Straight Line in Space exercise 27.5 question 6

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

Hint: using expression $\left|\frac{\mid \overrightarrow{a_{2}}-\overrightarrow{\left.a_{1}\right)} \times \overrightarrow{b_{1}}}{\vec{b}}\right|$

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

Solution:
Let line $x-1 / 2=\frac{y-2}{3}=\frac{z+y}{6}=\boldsymbol{\mu}$ ………… (i)
From above point $(x, y, z)$ on line 1 will be $(2 \boldsymbol{\mu}+1,3 \boldsymbol{\mu}+2,6 \boldsymbol{\mu}-4)$
Let line 2 $x-3 / 4=\frac{y-3}{6}=\frac{z+5}{12}=\lambda$ ...............(ii)
From above point $(x, y, z)$ on line 2 will be $(4 \lambda+3,6 \lambda+3,12 \lambda-5)$
Position vector from equation (i) we get
$\begin{aligned} \overrightarrow{8} &=(2 \mu+1) \hat{\imath}+(3 \mu+2) \hat{\jmath}+(6 \mu-4) \hat{k} \\\\ &=(\hat{\imath}+2 \hat{\jmath}-4 \hat{k})+(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \\\\ \overrightarrow{a_{1}} &=\hat{\imath}+2 \hat{\jmath}-4 \hat{k}, \overrightarrow{b_{1}}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \end{aligned}$
Position vector from evaluation (ii) we get
$\begin{aligned} \overrightarrow{8} &=(4 \lambda+3) \hat{\imath}+(6 \lambda+3) \hat{\jmath}+(12 \lambda-5) \hat{k} \\\\ &=(3 \hat{\imath}+3 \hat{\jmath}-5 \hat{k})+\lambda(4 \hat{\imath}+6 \hat{\jmath}+12 \hat{k}\\\\ \left.\overrightarrow{a_{2}}\right) &=3 \hat{\imath}+3 \hat{\jmath}-5 \hat{k}, \overrightarrow{b_{2}}=4 \hat{\imath}+6 \hat{\jmath}+12 \hat{k} \end{aligned}$
$\text { From } b_{1} \text { and } b_{2} \text { we get } \overrightarrow{b_{2}}=2 \overrightarrow{b_{1}}$
Shortest Distance $=\left|\frac{\left.\mid \overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \overrightarrow{b_{1}}}{\vec{b}}\right|$
$\left(\overrightarrow{a_{2}}-\overrightarrow{\left.a_{1}\right)}=(3 \hat{i}+3 \hat{j}-5 \hat{k})-(\hat{i}+2 \hat{\jmath}-4 \hat{k}=2 \hat{i}+\hat{j}-\hat{k}\right.$
$\left(\overrightarrow{a_{2}}-\overrightarrow{\left.a_{1}\right)} \; x\; \vec{b}\right.$ $=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{array}\right|=9 \hat{i}-14 \hat{\jmath}+4 \widehat{k}$
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{\left.a_{1}\right)} \times \vec{b}=\sqrt{19)^{2}+(-14)^{2}(4)^{2}}=\sqrt{81+196+16}=\sqrt{293}\right. \\\\ &|\vec{b}|=\sqrt{2^{2}+3^{2}+6^{2}=\sqrt{4+9+36}=7} \end{aligned}$
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 $\mathrm{D}=\left|\frac{\left.\overrightarrow{\left(a_{2}\right.}-\overrightarrow{\left.a_{1}\right)} \overrightarrow{\left(b_{1}\right.} \times \overrightarrow{b_{2}}\right)}{\left(b_{1} \times \overline{b_{2}}\right)}\right|$
Given: $\overrightarrow{8}=\hat{\imath}+2 \hat{\jmath}+\hat{k}+\lambda(\hat{\imath}-\hat{\jmath}+\hat{k}) \text { and } \overrightarrow{8}=2 \hat{\imath}-\hat{\jmath}-\hat{k}+\mu(2 \hat{\imath}+\hat{\jmath}+2 \hat{k})$
Solution:
$\begin{array}{ll} a_{1}=\hat{\imath}+2 \hat{\jmath}+\hat{k}, & b_{1}=\hat{\imath}-\hat{\jmath}-\hat{k} \\\\ a_{2}=2 \hat{\imath}-\hat{\jmath}-\hat{k} & b_{2}=2 \hat{\imath}+\hat{\jmath}+2 \hat{k} \\\\ \overrightarrow{a_{2}}-\overrightarrow{a_{1}}=\hat{\imath}-3 \hat{\jmath}-2 \hat{k} \end{array}$
$\overrightarrow{b_{1}}-\overrightarrow{b_{2}}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{array}\right|$
$\begin{aligned} &=\hat{\imath}(-3)-\hat{\jmath}(0)+\hat{k}(1+2) \\\\ &=-3 \hat{\imath}+3 \hat{k} \\\\ &\mathrm{SD}=\frac{(2 \hat{i}-\hat{j}-\hat{k})(-3 \hat{i}+3 \hat{k})}{\sqrt{9+9}} \end{aligned}$
$\begin{aligned} &=\left|\frac{6-3}{\sqrt{18}}\right| \\\\ &=\frac{9}{3 \sqrt{2}} \\\\ &=\frac{3}{\sqrt{2}} \quad \text { units } \end{aligned}$

Straight Line in Space exercise 27.5 question 7(ii)

Answer: $\mathrm{d}=2 \sqrt{29} \text { units }$
Hint: using expression $\left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|$
$\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}+a_{2} b_{1}\right)^{2}}$
Given: $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \quad \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$
Solution:
It is known that the shortest distance between the two lines,
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} \quad \text { and } \quad \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$
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
$\begin{aligned} &\left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|=\left|\begin{array}{ccc} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{array}\right| \\\\ &=4(-6+2)-6(7-1)+8(-14+6)=-116 \end{aligned}$
$\begin{aligned} &\sqrt{\left.b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{2}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}} \\\\ &=\sqrt{(-6+2)^{2}+(1+7)^{2}+(-14+6)^{2}} \\\\ &=\sqrt{16+36+64} \end{aligned}$
$\begin{aligned} &=\sqrt{116} \\\\ &=2 \sqrt{29} \end{aligned}$
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
$=2 \sqrt{29} \text { units }$

Straight Line in Space exercise 27.5 question 7(iii)

Answer: $d=\left|\frac{3}{\sqrt{19}}\right|$
Hint: using $\frac{\left.\overrightarrow{\left(b_{1}\right.} \times \overrightarrow{\left.b_{2}\right)} \overrightarrow{\left(a_{2}\right.}-\overrightarrow{a_{1}}\right)}{\left(b_{1} \times \overline{\left.b_{2}\right)}\right.}$
Given: $\overrightarrow{8}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}+\lambda(\hat{\imath}-3 \hat{\jmath}+2 \hat{k})$ and
$\overrightarrow{8}=4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k}+\mu(2 \hat{\imath}+3 \hat{\jmath}+\hat{k})$
Solution:
Shortest distance between the lines with vector equations
$\overrightarrow{8}=\overrightarrow{a_{1}}+\overrightarrow{b_{1}} \text { and } \overrightarrow{8}=\overrightarrow{a_{2}}+\overrightarrow{b_{2}} \text { is }\left|\frac{\overrightarrow{\left(b_{1}\right.} \times \overrightarrow{\left.b_{2}\right)}}{\left.\overline{\left(b_{1}\right.} \times \overline{a_{2}}-\overrightarrow{a_{1}}\right)}\right|$
Now
$\overrightarrow{8}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}+\lambda(\hat{\imath}-3 \hat{\jmath}+2 \hat{k})$ $\overrightarrow{8}=(4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k}+\mu(2 \hat{\imath}+3 \hat{\jmath}+\hat{k})$
Comparing with $\overrightarrow{8}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ Comparing with $\overrightarrow{8}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$
$\overrightarrow{a_{1}}=\hat{i}+2 \hat{j}+3 \widehat{k}$ $\overrightarrow{a_{2}}=4 \hat{i}+5 \hat{j}+6 \widehat{k}$
and and
$\overrightarrow{b_{1}}=\hat{i}-3 \hat{j}+2 \widehat{k}$ $\overrightarrow{b_{2}}=2 \hat{i}-3 \hat{j}+1 \widehat{k}$

$\begin{aligned} &\left.\overrightarrow{\left(a_{2}\right.}-\overrightarrow{a_{1}}\right)=(4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k}-(1 \hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\ &=(4-1) \hat{\imath}+(5-2) \hat{\jmath}+(6-3) \hat{k} \\ &=3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k} \end{aligned}$
$\overrightarrow{\left(b_{1}\right.} x \overrightarrow{\left.b_{2}\right)}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{array}\right|$
$\begin{aligned} &=\hat{\imath}[(-3 \times 1)-(3 \times 2)]-\hat{\jmath}[(1 \times 1)-(2 \times 2)]+\hat{k}[(1 \times 3)-(2 \times-3)] \\ &=\hat{\imath}(-9)-\hat{\jmath}(-3)+\hat{k}(9) \\ &=9 \hat{\imath}+3 \hat{\jmath}+9 \hat{k} \end{aligned}$
Magnitude of $\overrightarrow{\left(b_{1}\right.} \times \overrightarrow{\left.b_{2}\right)}=\sqrt{(-9)^{2}+3^{2}+9^{2}}$
$\begin{aligned} &=\sqrt{81+9+81} \\ &=\sqrt{171} \end{aligned}$
$\begin{aligned} &=\sqrt{9 \times 29} \\ &=3 \sqrt{19} \end{aligned}$
Also
$\begin{aligned} &\left.\overrightarrow{\left(b_{1}\right.} \times \overrightarrow{\left.b_{2}\right)} \overrightarrow{\left(a_{2}\right.}-\overrightarrow{a_{1}}\right)=(-9(+3 \hat{\jmath}+9 \hat{k})(3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}) \\ &=(-9 \times 3)+(3 \times 3)+(9+3) \\ &=-27+9+27 \\ &=9 \end{aligned}$
Shortest Distance $\begin{aligned} &=\left|\frac{9}{3 \sqrt{19}}\right| \\\\ \end{aligned}$
$=\left|\frac{3}{\sqrt{19}}\right|$

Straight Line in Space exercise 27.5 question 7(iv)

Answer: $d=9$
Hint: using the expression $d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}$
Given:

$\begin{aligned} &\overrightarrow{8}=6 \hat{\imath}+2 \hat{\jmath}+2 \hat{k}+\lambda(\hat{\imath}-2 \hat{\jmath}+2 \hat{k}) \text { and } \\\\ &\overrightarrow{8}=4 \hat{\imath}-\hat{k}+\boldsymbol{\mu}(2 \hat{\imath}+3 \hat{\jmath}+\hat{k}) \end{aligned}$

Solution:
$\begin{aligned} &\overrightarrow{a_{1}}=6 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \\\\ &\overrightarrow{a_{2}}=4 \hat{\imath}-\hat{k} \\\\ &\overrightarrow{b_{1}}=\hat{\imath}-2 \hat{\jmath}+2 \hat{k} \end{aligned}$
$\begin{aligned} &\overrightarrow{b_{2}}=\hat{3}_{l}-2 \hat{\jmath}-2 \hat{k} \\\\ &\overrightarrow{a_{2}}-\overrightarrow{a_{1}}=-10 \hat{\imath}-2 \hat{\jmath}+3 \hat{k} \end{aligned}$
$\text { and } \overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{array}\right| \Rightarrow 8 \hat{\imath}+8 \hat{\jmath}+4 \hat{k}$
$\begin{aligned} &\Rightarrow\left|\overrightarrow{b_{1}} \times \overline{b_{2}}\right| \Rightarrow \sqrt{8^{2}+8^{2}+4^{2}} \Rightarrow \sqrt{64+64+16} \\\\ &=\sqrt{144} \Rightarrow 12 \end{aligned}$
and
$\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=-10 \hat{\imath}-2 \hat{\jmath}+3 \hat{k}(8 \hat{\imath}+8 \hat{\jmath}+4 \hat{k}) \\\\ &=-80-16-12 \\\\ &=-108 \end{aligned}$
The shortest distance between the lines
$\begin{aligned} &\overrightarrow{8}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \text { and } \overrightarrow{8}=\overrightarrow{a_{2}}+\mu b_{2} \\\\ &\mathrm{~d}=\left|\frac{-108}{12}\right| \\\\ &=9 \end{aligned}$

Straight Line in Space exercise 27.5 question 8

Answer: $D=\frac{1}{7}$
Hint: using formula distance between $\parallel$ line $\left|\frac{\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \vec{b}}{\vec{b}}\right|$
Given: $\overrightarrow{8}=\hat{\imath}+2 \hat{\jmath}-4 \hat{k}++\lambda(2 \hat{\imath}-3 \hat{\jmath}+6 \hat{k}) \text { and }$
$\overrightarrow{8}=3 \hat{\imath}+3 \hat{\jmath}-5 \hat{k}+\mu(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k})$
Solution:
By the given formula,
$\begin{aligned} &=\left|\frac{(2 \hat{\imath}+\hat{\jmath}-\hat{k})(2 \hat{i}+3 \hat{j}+6 \hat{k})}{\sqrt{4+9+36}}\right| \\\\ &=\frac{4+3-6}{\sqrt{49}} \end{aligned}$
$\begin{aligned} &=\frac{4+3-6}{7} \\\\ &=1 / 7 \end{aligned}$

Straight Line in Space exercise 27.5 question 9

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

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.

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