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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
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The correct answer is:
$\left(\frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)$
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
$\begin{aligned} & \left(\frac{2}{\sqrt{2^2+3^2+5^2}}, \frac{-3}{\sqrt{2^2+3^2+5^2}}, \frac{5}{\sqrt{2^2+3^2+5^2}}\right) \\ & =\left(\frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)\end{aligned}$
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
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
$\begin{aligned} & \left(\frac{2}{\sqrt{2^2+3^2+5^2}}, \frac{-3}{\sqrt{2^2+3^2+5^2}}, \frac{5}{\sqrt{2^2+3^2+5^2}}\right) \\ & =\left(\frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)\end{aligned}$
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