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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 6 maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 6 maths textbook solution

Answers (1)

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

 

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