A plane makes positive intercepts of unit len | vedclass

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(D) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Given $a = 1$ and $b = 1$,the equation becomes $x

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{2}{3}\right)$

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(D) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Given $a = 1$ and $b = 1$,the equation becomes $x + y + \frac{z}{c} = 1$. Since the plane passes through $(-1, 1, 2)$,we substitute these coordinates: $-1 + 1 + \frac{2}{c} = 1$,which gives $\frac{2}{c} = 1$,so $c = 2$. The equation of the plane is $x + y + \frac{z}{2} = 1$,or $2x + 2y + z - 2 = 0$. The normal vector to the plane is $\vec{n} = 2\hat{i} + 2\hat{j} + 1\hat{k}$. The direction vector of the $X$-axis is $\vec{v} = 1\hat{i} + 0\hat{j} + 0\hat{k}$. The angle $	heta$ between a line with direction vector $\vec{v}$ and a plane with normal $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{n} \cdot \vec{v}|}{|\vec{n}| |\vec{v}|}$. $\sin 	heta = \frac{|(2)(1) + (2)(0) + (1)(0)|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{1^2 + 0^2 + 0^2}} = \frac{2}{\sqrt{4 + 4 + 1} \cdot 1} = \frac{2}{\sqrt{9}} = \frac{2}{3}$. Thus,$	heta = \sin^{-1}\left(\frac{2}{3}ight)$.

$A$ plane makes positive intercepts of unit length on each of $X$ and $Y$ axes. If it passes through the point $(-1, 1, 2)$ and makes an angle $\theta$ with the $X$-axis,then $\theta$ is

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