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

The direction ratios of the line are $1,2,k$.

The direction ratios of the normal to the plane are $2,-3,3$.

Let $\theta$ be the angle between the line and the plane. The sine of the angle is given by:

$\sin \theta =\frac{|1(2)+2(-3)+k(3)|}{\sqrt{1^2+2^2+k^2}\sqrt{2^2+(-3{)}^2+3^2}}=\frac{|3k-4|}{\sqrt{5+k^2}\sqrt{22}}$

Given $\theta ={\cos }^{-1}\left(\sqrt{\frac{13}{22}}\right)$, we have $\cos \theta =\sqrt{\frac{13}{22}}$.

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{13}{22}}=\sqrt{\frac{9}{22}}=\frac{3}{\sqrt{22}}$

Equating the two expressions for $\sin \theta$:

$\frac{|3k-4|}{\sqrt{5+k^2}\sqrt{22}}=\frac{3}{\sqrt{22}}$

$|3k-4|=3\sqrt{5+k^2}$

Squaring both sides:

$(3k-4{)}^2=9(5+k^2)$

$9k^2-24k+16=45+9k^2$

$-24k=29$

$k=-\frac{29}{24}$

Answer: $-\frac{29}{24}$