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Question

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


Solution

The direction of ratios of the lines $$\cfrac{x-1}{-3}=\cfrac{y-2}{2k}=\cfrac{z-3}{2}$$ and $$\cfrac{x-1}{3k}=\cfrac{y-1}{1}=\cfrac{z-6}{-5}$$ are $$-3, 2k, 2$$ and $$3k, 1, -5$$ respectively.
It is known that two lines with direction ratios $${a}_{1}, {b}_{1}, {c}_{1}$$ and $${a}_{2}, {b}_{2},
{c}_{2}$$ are perpendicular, if $${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0$$
$$\therefore$$ $$-3(3k)+2k\times 1+2(-5)=0$$
$$\Rightarrow$$ $$-9k+2k-10=0$$
$$\Rightarrow$$ $$7k=-10$$
$$\Rightarrow$$ $$k=\cfrac{-10}{7}$$
Therefore for $$k=-\cfrac{10}{7}$$, the given lines are perpendicular to each other.

Mathematics
RS Agarwal
Standard XII

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