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Straight line in space — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSStraight line in space3 Marks
Question
It the lines $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{2\lambda}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}-6}{-5}$ are perpendicular, find the value of $\lambda.$
✓

Answer

The diraction of ratios of the lines, $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{2\lambda}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}-6}{-5},$ are $-3, 2k, 2$ and $3k, 1, -5$ respectiveiy.
It is know that two lines with direction ratios $, a_1, b_1, c_1$ and $a_2, b_2, c_2,$ are perpendicular, if $a_{1 }a_{2 }+ b_{1 }b_{2 }+ c_{1 }c_{2 }= 0$
$\therefore -3 (3k) + 2k \times 1 + 2 (-5) = 0$
$\Rightarrow -9k + 2k - 10 = 0$
$\Rightarrow 7k = - 10$
$\Rightarrow\text{k}=\frac{-10}{7}$
Therefore, for $\text{k}=-\frac{10}{7},$ the given lines are perpendicular to each other.

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