Question

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

• A. \(\dfrac43\)
• B. \(\dfrac53\)
• C. \(\dfrac23\)
• D. \(\dfrac73\)

Hint:

For a line and a plane, \[ \sin\theta= \frac{|\,\text{direction vector}\cdot\text{normal vector}\,|} {|\text{direction vector}|\,|\text{normal vector}|}. \] This formula appears very frequently in three-dimensional geometry.

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 satisfies \[ \sin\theta = \frac{|al+bm+cn|} {\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}. \] This relation follows from the fact that the angle between a line and a plane is complementary to the angle between the line and the plane's normal.

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

Step 3: Apply the formula for angle between a line and a plane. \[ \sin\theta = \frac{|2(1)+(-1)(2)+2\sqrt{\lambda}|} {\sqrt{4+1+\lambda}\sqrt{1+4+4}}. \] \[ = \frac{|2-2+2\sqrt{\lambda}|} {3\sqrt{5+\lambda}}. \] \[ = \frac{2\sqrt{\lambda}} {3\sqrt{5+\lambda}}. \]

Step 4: Use the given value of \(\sin\theta\). Given \[ \sin\theta=\frac13. \] Therefore, \[ \frac{2\sqrt{\lambda}} {3\sqrt{5+\lambda}} = \frac13. \] Multiplying both sides by \(3\), \[ \frac{2\sqrt{\lambda}} {\sqrt{5+\lambda}} = 1. \] Squaring both sides, \[ \frac{4\lambda} {5+\lambda} = 1. \] \[ 4\lambda = 5+\lambda. \] \[ 3\lambda=5. \] \[ \lambda=\frac53. \]

Step 5: Final Conclusion. \[ \boxed{\lambda=\frac53} \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]