A plane makes positive intercepts of unit length on each of $X$ and $Y$ axis. If it passes through the point…

A plane makes positive intercepts of unit length on each of $X$ and $Y$ axis. If it passes through the point $(-1,1,2)$ and makes angle $\theta$ with the X -axis, then $\theta$ is
  1. $\cos ^{-1}\left(\frac{2}{3}\right)$
  2. $\cos ^{-1}\left(\frac{1}{3}\right)$
  3. $\sin ^{-1}\left(\frac{1}{3}\right)$
  4. $\sin ^{-1}\left(\frac{2}{3}\right)$

Solution

Equation of plane in intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Here, $\mathrm{a}=1, \mathrm{~b}=1$ $\therefore \quad \frac{x}{1}+\frac{y}{1}+\frac{z}{c}=1$
Since this plane passes through the point $(-1,1,2)$ $\begin{aligned} \therefore \quad & -1+1+\frac{2}{c}=1 \\ & \Rightarrow c=2 \end{aligned}$ $\therefore \quad$ Equation of plane is $\begin{aligned} & x+y+\frac{z}{2}=1 \\ & \Rightarrow 2 x+2 y+z=2 \end{aligned}$ D.r.s of X -axis are $1,0,0$. $\begin{aligned} & \therefore \quad \sin \theta=\frac{2(1)+0+0}{\sqrt{4+4+1} \sqrt{1}} \\ & \quad \Rightarrow \sin \theta=\frac{2}{3} \\ & \quad \Rightarrow \theta=\sin ^{-1}\left(\frac{2}{3}\right) \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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