class 12 maths three dimensional geometry

If the fines $\cfrac{{x - 1}}{{ - 3}} = \cfrac{{y - 2}}{{2k}} = \cfrac{{z - 3}}{2}$
and $\cfrac{{x - 1}}{{3k}} = \cfrac{{y - 1}}{1} = \cfrac{{z - 6}}{{ - 5}}$ are perpendicular, then find the value of k.

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📘 Three Dimensional Geometry NCERT,Misc,Q.6, Page .498 SA

If the fines $\cfrac{{x - 1}}{{ - 3}} = \cfrac{{y - 2}}{{2k}} = \cfrac{{z - 3}}{2}$
and $\cfrac{{x - 1}}{{3k}} = \cfrac{{y - 1}}{1} = \cfrac{{z - 6}}{{ - 5}}$ are perpendicular, then find the value of k.

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

The given lines are $\cfrac{{x - 1}}{{ - 3}} = \cfrac{{y - 2}}{{2k}} = \cfrac{{z - 3}}{2}$ ...(1)

and $\cfrac{{x - 1}}{{3k}} = \cfrac{{y - 1}}{1} = \cfrac{{z - 6}}{{ - 5}}$ ...(2)

The direction ratios of line (1) are $< - 3,2k,2 >$

The direction ratios of line (2) are $< 3k,1, - 5 >$

The line (1) and (2) are perpendicular,

So, $( - 3)(3k) + 2(k)(1) + (2)( - 5) = 0$

$\Rightarrow$ $- 9k + 2k - 10 = 0 \Rightarrow 7k = - 10 \Rightarrow k = - \cfrac{{10}}{7}$

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