The value of k for which the lines x - 1-3 - y - 22k - z-32 and x-13k - y - 51 - z-67 may be perpendicular is-

Given : nbsp;L_1:x + 1 3 = y + 22k = z 32 D.R's of line L_1=( 3,2k,2) L_2:x 13k = y + 51 = z+67 D.R's of line L_2=(3k,1,7) we know, angle between two lines c ...

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Byju's Answer

Standard XII

Mathematics

Distinguish Acute Angle Bisectors and Obtuse Angle Bisectors

The value of ...

Question

The value of $k$ for which the lines $\frac{x + 1}{- 3} = \frac{y + 2}{2 k} = \frac{z - 3}{2} and \frac{x - 1}{3 k} = \frac{y + 5}{1} = \frac{z + 6}{7}$ may be perpendicular is:

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Solution

The correct option is B $2$
Given : $L_{1} : \frac{x + 1}{- 3} = \frac{y + 2}{2 k} = \frac{z - 3}{2}$
$D . R^{'} s$ of line $L_{1} = (- 3, 2 k, 2)$
$L_{2} : \frac{x - 1}{3 k} = \frac{y + 5}{1} = \frac{z + 6}{7}$
$D . R^{'} s$ of line $L_{2} = (3 k, 1, 7)$
we know, angle between two lines $cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
For lines to be perpendicular, $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ $\Rightarrow (- 3) (3 k) + (2 k) (1) + 2 (7) = 0 \Rightarrow k = 2$

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Similar questions

Q. If the lines

\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}

and

\frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}

are perpendicular, find the value of

k

If the lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} a n d \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ perpendicular, then find the value of k.