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  • Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 9

Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 9

Answers (1)

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 }

 

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