Question

Find the angle between the line \[ \frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6} \] and the plane \(10x+2y-11z=3\).

• A. \(\sin^{-1}\!\left(\dfrac{8}{21}\right)\)
• B. \(\sin^{-1}\!\left(\dfrac{5}{21}\right)\)
• C. \(\sin^{-1}\!\left(\dfrac{7}{21}\right)\)
• D. \(\sin^{-1}\!\left(\dfrac{1}{21}\right)\)

Hint:

Angle between a line and plane uses sine with direction ratios and plane normal.

Answer 1

The Correct Option is A

Step 1: Direction ratios of the line are \((2,3,6)\).
Step 2: Normal vector of the plane is \((10,2,-11)\).
Step 3: Angle \(\theta\) between line and plane: \[ \sin\theta=\frac{|2\cdot10+3\cdot2+6(-11)|}{\sqrt{(2^2+3^2+6^2)(10^2+2^2+11^2)}} =\frac{8}{21} \]