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  • The shortest distance between the lines<img alt="\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2} \text { and } \frac{x-6}{3}=\frac{1-y}{2}

The shortest distance between the lines\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2} \text { and } \frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}  is equal to

Option: 1

14


Option: 2

--


Option: 3

--


Option: 4

--


Answers (1)

\begin{aligned} \mathrm{L}_1: \overline{\mathrm{a}} & =<2,-1,6> & \mathrm{L}_2: \overline{\mathrm{b}} & =<6,1,-8> \\ \overline{\mathrm{p}} & =<3,2,2> & \overline{\mathrm{q}} & =<3,-2,0> \end{aligned}

\overline{\mathrm{p}} \times \overline{\mathrm{q}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{array}\right|\begin{align*} &= \langle 4,6,-12> \\ &= \langle 2,3,-6> \end{align*} (s.f)

\begin{aligned} & \mathrm{b} \Delta=\left|\frac{(\overline{\mathrm{b}}-\overline{\mathrm{a}}) \cdot|\overline{\mathrm{p}} \times \overline{\mathrm{q}}|}{|\overline{\mathrm{p}} \times \overline{\mathrm{q}}|}\right| \\ & =\left|\frac{(4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-14 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})}{\sqrt{4+9+36}}\right| \\ & =\left|\frac{8+6+84}{\sqrt{40}}\right|=\left|\frac{98}{7}\right|=14 \end{aligned}

Posted by

Ramraj Saini

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