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Shortest distance between lines \frac{x-5}{2}=\frac{y+19}{-9}=\frac{z+1}{-2} and \frac{x+3}{2}=\frac{y+12}{-9}=\frac{z-4}{-2} is

  • Option 1)

    6

  • Option 2)

    7

  • Option 3)

    8

  • Option 4)

    9

 

Answers (1)

best_answer

As we have learned

Shortest distance between two parallel lines (Cartesian form) -

Shortest distance between \frac{(x-x_{1})}{a}= \frac{(y-y_{1})}{b}= \frac{(z-z_{1})}{c}and

\frac{(x-x_{2})}{a}= \frac{(y-y_{2})}{b}= \frac{(z-z_{2})}{c} is given by

\frac{\begin{Vmatrix} \hat{i} &\hat{j} &\hat{k} \\ x_{2}-x_{1}&y_{2}-y_{1} &z_{2}-z_{1} \\ a& b & c \end{Vmatrix}}{\sqrt{a^{2}+b^{2}+c^{2}}}

 

-

 

 Since lines are parallel so formula for distance between parallel lines will be used 

Here, a=2, b=-9, c=-2

(x_1,y_1,z_1)=(5,-19,-1) and (x_2,y_2,z_2)=(-3,-12,4)

\therefore\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ x_2-x_1 &y_2-y_1 &z_2-z_1 \\ a &b & c \end{vmatrix}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ -8& 7 &5 \\ 2&-9 &-2 \end{vmatrix}=\hat{i}(31)-\hat{j}(6)+\hat{k}(58)=31\hat{i}-6\hat{j}+58\hat{k}\therefore Shortest distance \frac{|31\hat{i}-6\hat{j}+58\hat{k}|}{\sqrt{4+81+4}}=\sqrt{49}=7

 

 

 


Option 1)

6

Option 2)

7

Option 3)

8

Option 4)

9

Posted by

Himanshu

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