Question

If the angle \(\theta\) between the line \[ \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2} \] and the plane \[ 2x-y+\sqrt{\lambda}z+4=0 \] is such that \(\sin\theta=\frac13\), then the value of \(\lambda\) is:

• A. \(\frac53\)
• B. \(-\frac35\)
• C. \(-\frac34\)
• D. \(-\frac43\)

Hint:

For angle between 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 A

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 satisfies: \[ \sin\theta = \frac{|al+bm+cn|} {\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}} \]

Step 1: Extract the direction ratios of the line.
From: \[ \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2} \] the direction ratios are: \[ (1,2,2) \]

Step 2: Find the normal vector of the plane.
The plane is: \[ 2x-y+\sqrt{\lambda}z+4=0 \] Hence the normal vector is: \[ (2,-1,\sqrt{\lambda}) \]

Step 3: Use the angle formula.
Given: \[ \sin\theta=\frac13 \] Thus, \[ \frac{|2(1)+(-1)(2)+\sqrt{\lambda}(2)|} {\sqrt{4+1+\lambda}\sqrt{1+4+4}} = \frac13 \] Simplifying numerator: \[ 2-2+2\sqrt{\lambda} = 2\sqrt{\lambda} \] Denominator: \[ \sqrt{\lambda+5}\cdot3 \] Hence: \[ \frac{2\sqrt{\lambda}} {3\sqrt{\lambda+5}} = \frac13 \]

Step 4: Solve for \(\lambda\).
Cross multiplying: \[ 2\sqrt{\lambda} = \sqrt{\lambda+5} \] Squaring both sides: \[ 4\lambda=\lambda+5 \] \[ 3\lambda=5 \] \[ \lambda=\frac53 \] Hence, \[ \boxed{\frac53} \]