If the angel $$\theta $$ 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 $$\sin \,\theta = \frac{1}{3}$$ then the value of $$\lambda $$ is :
A.
$$\frac{5}{3}$$
B.
$$\frac{{ - 3}}{5}$$
C.
$$\frac{3}{4}$$
D.
$$\frac{{ - 4}}{3}$$
Answer :
$$\frac{5}{3}$$
Solution :
If $$\theta $$ is the angle between line and plane then $$\left( {\frac{\pi }{2} - \theta } \right)$$ is the angle between line and normal to plane given by
$$\eqalign{
& \cos \,\left( {\frac{\pi }{2} - \theta } \right) = \frac{{\left( {\hat i + 2\hat j + 2\hat k} \right).\left( {2\hat i - \hat j + \sqrt \lambda \hat k} \right)}}{{3\sqrt {4 + 1 + \lambda } }} \cr
& \cos \left( {\frac{\pi }{2} - \theta } \right) = \frac{{2 - 2 + 2\sqrt \lambda }}{{3 \times \sqrt 5 + \lambda }} \cr
& \Rightarrow \sin \,\theta = \frac{{2\sqrt \lambda }}{{3\sqrt 5 + \lambda }} = \frac{1}{3} \cr
& \Rightarrow 4\lambda = 5 + \lambda \cr
& \Rightarrow \lambda = \frac{5}{3} \cr} $$
Releted MCQ Question on Geometry >> Three Dimensional Geometry
Releted Question 1
The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :
If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :