Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the equation of the plane in intercept form be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

Given that the plane makes positive intercepts of unit length on the X and Y axes, we have $a=1$ and $b=1$.

The equation of the plane becomes:
$x+y+\frac{z}{c}=1$

Since the plane passes through the point $(-1,1,2)$, we substitute these coordinates into the equation:
$-1+1+\frac{2}{c}=1$

$\frac{2}{c}=1\Rightarrow c=2$

Substituting $c=2$ back into the equation of the plane, we get:
$x+y+\frac{z}{2}=1$

$2x+2y+z=2$

The normal vector to the plane is $\vec{n}=2\hat{i}+2\hat{j}+\hat{k}$.
The direction vector of the X-axis is $\vec{u}=\hat{i}$.

The angle $\theta$ between a plane and a line is given by:
$\sin \theta =\frac{|\vec{n}\cdot \vec{u}|}{|\vec{n}||\vec{u}|}$

Calculating the dot product and magnitudes:
$\vec{n}\cdot \vec{u}=(2)(1)+(2)(0)+(1)(0)=2$
$|\vec{n}|=\sqrt{2^2+2^2+1^2}=\sqrt{4+4+1}=3$
$|\vec{u}|=\sqrt{1^2+0^2+0^2}=1$

Therefore,
$\sin \theta =\frac{2}{3\times 1}=\frac{2}{3}$

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

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