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

Step 1: Recall the line and plane angle.
If a line has direction ratios $(l,m,n)$ and a plane has normal $(a,b,c)$, the angle $\theta$ between them obeys \[ \sin\theta=\frac{|al+bm+cn|}{\sqrt{a^2+b^2+c^2}\,\sqrt{l^2+m^2+n^2}}. \]
Step 2: Read the direction ratios of the line.
From $\dfrac{x+1}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{2}$ the ratios are $(1,2,2)$.
Step 3: Read the normal of the plane.
The plane $2x-y+\sqrt{\lambda}\,z+4=0$ has normal $(2,-1,\sqrt{\lambda})$.
Step 4: Plug into the formula.
Numerator: $|2(1)+(-1)(2)+\sqrt{\lambda}(2)|=|2\sqrt{\lambda}|$. Denominator: $\sqrt{4+1+\lambda}\cdot\sqrt{1+4+4}=3\sqrt{\lambda+5}$. With $\sin\theta=\dfrac13$, \[ \frac{2\sqrt{\lambda}}{3\sqrt{\lambda+5}}=\frac13. \]
Step 5: Cross multiply and simplify.
$2\sqrt{\lambda}=\sqrt{\lambda+5}$.
Step 6: Square and solve.
$4\lambda=\lambda+5$, so $3\lambda=5$ and $\lambda=\dfrac53$. \[ \boxed{\frac53} \]