Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The direction vector of the given line is $\vec{d}=\hat{i}+\hat{j}+\hat{k}$.

The normal vector to the given plane is $\vec{n}=2\hat{i}-\hat{j}+\hat{k}$.

Let $\theta$ be the angle between the line and the plane. Then $\sin \theta$ is given by:

$\sin \theta =\frac{|\vec{d}\cdot \vec{n}|}{|\vec{d}||\vec{n}|}$

$\sin \theta =\frac{|(1)(2)+(1)(-1)+(1)(1)|}{\sqrt{1^2+1^2+1^2}\sqrt{2^2+(-1{)}^2+1^2}}=\frac{2}{\sqrt{3}\sqrt{6}}=\frac{2}{3\sqrt{2}}=\frac{\sqrt{2}}{3}$

In $△ABC$, $AB$ is perpendicular to the plane, so $\angle B=90^{\circ }$. The angle at $C$ is the angle between the line and the plane, which is $\theta$.

Using trigonometric ratios in $△ABC$:

$\sin \theta =\frac{AB}{AC}$

Substituting the known values:

$\frac{\sqrt{2}}{3}=\frac{4}{AC}$

$AC=\frac{12}{\sqrt{2}}=6\sqrt{2}$

Squaring both sides gives $A{C}^2=72$.

Using Pythagoras theorem in $△ABC$:

$A{C}^2=A{B}^2+B{C}^2$

$72=4^2+B{C}^2$

$B{C}^2=72-16=56$

Answer: $56$