If $-2, \frac{4}{3}, \frac{-4}{5}$ are the intercepts made by a plane on $\mathrm{X}, \mathrm{Y},…

If $-2, \frac{4}{3}, \frac{-4}{5}$ are the intercepts made by a plane on $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$-axes respectively then the direction cosines of a normal to this plane are
  1. $\left(\frac{-1}{3}, \frac{2}{3}, \frac{-2}{3}\right)$
  2. $\left(\frac{2}{3 \sqrt{5}}, \frac{-4}{3 \sqrt{5}}, \frac{5}{3 \sqrt{5}}\right)$
  3. $\left(\frac{-4}{\sqrt{57}}, \frac{4}{\sqrt{57}}, \frac{-5}{\sqrt{57}}\right)$
  4. $\left(\frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)$

Solution

If plane makes intercepts $a, b, c$ then equation of plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ for $-2,4 / 3,-4 / 5$, equation of plane is $\begin{aligned} & \frac{x}{-2}+\frac{y}{\left(\frac{4}{3}\right)}=\frac{z}{\left(\frac{-4}{5}\right)}=1 \\ & \frac{x}{-2}+\frac{3 y}{4}-\frac{52}{4}=1 \\ & \frac{-2 x+3 y-52}{4}=1\end{aligned}$ $\begin{aligned} & -2 x+3 y-52=4 \\ & 2 x-3 y+52+4=0\end{aligned}$ Direction ratio of normal to plane are $(2,-3,5)$ Direction ratio normal to plane are $\begin{aligned} & \left(\frac{2}{\sqrt{2^2+3^2+5^2}}, \frac{-3}{\sqrt{2^2+3^2+5^2}}, \frac{5}{\sqrt{2^2+3^2+5^2}}\right) \\ & =\left(\frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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