Question

If the angle between the line \(x=\frac{y-5}{2}=\frac{z+8}{k}\) and the plane \(2x-3y+3z=9\) is \(\cos ^{-1}\left(\sqrt{\frac{13}{22}}\right)\), then \(k=\ldots\)

• A. \(-\frac{24}{29}\)
• B. \(\frac{24}{29}\)
• C. \(-\frac{29}{24}\)
• D. \(\frac{29}{24}\)

Hint:

The sine of the angle between a line and a plane is |d . n| / (|d||n|); convert the given cosine into a sine.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Line direction \(\bar{d}=(1,2,k)\), plane normal \(\bar{n}=(2,-3,3)\). The angle \(\theta\) with the plane satisfies \(\sin \theta =\frac{|\bar{d}\cdot \bar{n}|}{|\bar{d}||\bar{n}|}\).

Step 2: Convert the given angle
\(\cos \theta =\sqrt{\frac{13}{22}}\Rightarrow \sin \theta =\sqrt{\frac{9}{22}}=\frac{3}{\sqrt{22}}\).

Step 3: Set up
\(\bar{d}\cdot \bar{n}=3k-4\), \(|\bar{n}|=\sqrt{22}\), so \(\frac{|3k-4|}{\sqrt{5+k^{2}}}=3\).

Step 4: Solve
\(9k^{2}-24k+16=45+9k^{2}\Rightarrow -24k=29\Rightarrow k=-\frac{29}{24}\).

Final Answer:
The value of k is \(-\frac{29}{24}\). \[ \boxed{\text{(C) }-\dfrac{29}{24}} \]