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The Direction Ratios of the Line Perpendicular to the Lines X − 7 2 = Y + 17 − 3 = Z − 6 1 a N D , X + 5 1 = Y + 3 2 = Z − 4 − 2 (A) 4, 5, 7 (B) 4, −5, 7 (C) 4, −5, −7 (D) −4, 5, 7

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Question

The direction ratios of the line perpendicular to the lines \[\frac{x - 7}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1} \text{ and }, \frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}\] are proportional to

Options

  • (a) 4, 5, 7

  • (b) 4, −5, 7

  • (c) 4, −5, −7

  • (d) −4, 5, 7

     
MCQ
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Solution

(a) 4, 5, 7 

We have ,

\[\frac{x - 7}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1} \]

\[\frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}\]

The direction ratios of the given lines are proportional to 2,-3,1 and 1,2,-2.

The vectors parallel to the given vectors are \[\overrightarrow{b_1} = 2 \hat{i} - 3 \hat{j} + \hat{k}  \text{ and  } \overrightarrow{b_2} = \hat{i}  + 2 \hat{j}  - 2 \hat{k} \]

Vector perpendicular to the given two lines is ,

\[\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b}_2 \]

\[ = \begin{vmatrix}\hat{i} & \hat{j}  & \hat{k}  \\ 2 & - 3 & 1 \\ 1 & 2 & - 2\end{vmatrix}\]

\[ = 4 \hat{i}  + 5 \hat{j}  + 7 \hat{k} \] 

Hence, the direction ratios of the line perpendicular to the given two lines are proportional to 4, 5, 7.

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