The direction cosines of the line which is perpendicular to the lines…

The direction cosines of the line which is perpendicular to the lines $\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}$ and $\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}$ are
  1. $\pm \frac{3}{\sqrt{50}}, \pm \frac{4}{\sqrt{50}}, \pm \frac{5}{\sqrt{50}}$
  2. $\pm \frac{4}{\sqrt{90}}, \pm \frac{5}{\sqrt{90}}, \pm \frac{7}{\sqrt{90}}$
  3. $\pm \frac{4}{\sqrt{29}}, \pm \frac{3}{\sqrt{29}}, \pm \frac{2}{\sqrt{29}}$
  4. $\pm \frac{1}{\sqrt{26}}, \pm \frac{3}{\sqrt{26}}, \pm \frac{4}{\sqrt{26}}$

Solution

D.R.'s can be obtained by $\begin{aligned} & \frac{a}{(-3)(-2)-2 \times 1}=\frac{b}{1 \times 1-2 \times(-2)}=\frac{c}{2 \times 2-2 \times(-2)} \\ & \Rightarrow \frac{a}{4}=\frac{b}{5}=\frac{c}{7}\end{aligned}$ So, direction cosines are $\pm \frac{4}{\sqrt{4^2+5^2+7^2}}, \pm \frac{5}{\sqrt{4^2+5^2+7^2}}, \pm \frac{4}{\sqrt{4^2+5^2+7^2}}$ i.e., $\pm \frac{4}{\sqrt{90}}, \pm \frac{5}{\sqrt{90}}, \pm \frac{7}{\sqrt{90}}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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