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  • The shortest distance between the lines   <img alt="\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx&plus;2%7D%7B1%7D%3D%5Cfr

The shortest distance between the lines   \frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}    and   \frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}     is:

 

Option: 1

8


Option: 2

7


Option: 3

6


Option: 4

9


Answers (1)

best_answer

\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}     and \frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}

Shortest distance (9) =\frac{\left\|\begin{array}{ccc} \mathrm{a}_2-\mathrm{a}_1 & \mathrm{~b}_2-\mathrm{b}_1 & \mathrm{c}_2-\mathrm{c}_1 \\ \mathrm{l}_1 & \mathrm{~m}_1 & \mathrm{n}_1 \\ \mathrm{l}_2 & \mathrm{~m}_2 & \mathrm{n}_2 \end{array}\right\|}{\left\|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \mathrm{k} \\ \mathrm{l}_1 & \mathrm{~m}_1 & \mathrm{n}_1 \\ \mathrm{l}_2 & \mathrm{~m}_2 & \mathrm{n}_2 \end{array}\right\|}

=\begin{bmatrix} 4+2 &1-0 &-3-5 \\ 1 &-2 &2 \\ 1& 2& 0 \end{bmatrix}

=\begin{bmatrix} i\hat{}&j\hat{} &-k \\ 1 &-2 &2 \\ 1& 2& 0 \end{bmatrix}

=\begin{bmatrix} 6&1 &-8 \\ 1 &-2 &2 \\ 1& 2& 0 \end{bmatrix}

\begin{aligned} & =\frac{}{|\hat{i}(-4)-\hat{j}(-2)+\mathrm{k}(2+2)|} \\ & =\frac{|-54|}{|-4 \hat{i}+2 \hat{j}+4 k|} \\ & =\frac{54}{\sqrt{16+4+16}} \\ & =\frac{54}{6} \\ \end{aligned}

=9

 

Posted by

Anam Khan

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