Question

An angle between the plane \[ x+y+z=5 \] and the line \[ \frac{x-16}{0} = \frac{y-0}{-1} = \frac{z+47}{4} \] is

• A. \[ \sin^{-1}\!\left(\frac{3}{\sqrt{17}}\right) \]
• B. \[ \sin^{-1}\!\left(\sqrt{\frac{3}{17}}\right) \]
• C. \[ \cos^{-1}\!\left(\sqrt{\frac{3}{17}}\right) \]
• D. \[ \sin^{-1}\!\left(\frac{5}{\sqrt{17}}\right) \]

Hint:

For a line and a plane: \[ \sin\theta= \frac{|\vec n\cdot\vec d|} {|\vec n|\,|\vec d|}, \] where \(\vec n\) is the normal vector of the plane and \(\vec d\) is the direction vector of the line.

Answer 1

The Correct Option is B

Concept: If a line has direction ratios \[ (l,m,n) \] and a plane has normal vector \[ (a,b,c), \] then the angle \(\theta\) between the line and the plane is given by \[ \sin\theta = \frac{|al+bm+cn|} {\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}. \]

Step 1: Find the direction ratios of the line. Given \[ \frac{x-16}{0} = \frac{y}{-1} = \frac{z+47}{4}. \] Hence the direction ratios of the line are \[ (0,-1,4). \]

Step 2: Find the normal vector of the plane. The plane is \[ x+y+z=5. \] Its normal vector is \[ (1,1,1). \]

Step 3: Apply the formula for the angle between a line and a plane. Using \[ (a,b,c)=(1,1,1), \] and \[ (l,m,n)=(0,-1,4), \] we get \[ \sin\theta = \frac{|1(0)+1(-1)+1(4)|} {\sqrt{1^2+1^2+1^2}\sqrt{0^2+(-1)^2+4^2}}. \] \[ = \frac{3}{\sqrt3\sqrt{17}}. \] \[ = \frac{\sqrt3}{\sqrt{17}}. \] \[ = \sqrt{\frac{3}{17}}. \] Therefore, \[ \theta = \sin^{-1}\!\left(\sqrt{\frac{3}{17}}\right). \]

Step 4: Write the final answer. \[ \boxed{ \sin^{-1}\!\left(\sqrt{\frac{3}{17}}\right) } \]