Consider the line L : $\vec{r} + (\hat{i} + 3\hat{j} - \hat{k})+ λ(\hat{i} + 2\hat{k}) $ and the plane $\pi : \vec{r} .(\hat{i} + 4\hat{j} + \hat{k}) + 6 = 0 .$
Statement-1: The line L intersects the plane $\pi $ at the point (1 , 0, -7).
Statement-2: The angle θ between the line L and the plane $\pi $ is given by $\theta = \frac{1}{2} cos^{-1} (\frac{1}{5})$.
Answer & explanation
Correct answer: option 4
The position vector of any point on the line is $\hat{i} + ( 3 + λ )\hat{j} + (2 λ - 1) \hat{k}$.
If this point lies on the given plane.
$\begin{Bmatrix}\hat{i} + ( 3 + λ )\hat{j} + (2 λ - 1) \hat{k}\end{Bmatrix}. \begin{Bmatrix}\hat{i} + 4 \hat{j} + \hat{k}\end{Bmatrix} + 6 = 0 $
$⇒ 1 + 12 + 4λ + 2λ - 1 + 6 = 0 ⇒ 6λ + 18 = 0 ⇒λ = - 3$
So, the position vector of the point of intersection is $\hat{i} - 7\hat{k}.$
so $x=(1+λ)$
$y=3$
$z=-1+2λ$ at point (1, 0, -7)
$x=1+λ=1⇒λ=0$
$y=3⇒y=0$ not possible
So, statement-1 is false.
The angle θ between the given line and given plane is given by
$sin \theta = \frac{(\hat{i} + 2\hat{k}).(\hat{i} + 4\hat{j}+\hat{k})}{\sqrt{5}\sqrt{18}}$ $[∵ sin \theta = \frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|}]$
$⇒ sin \theta = \sqrt{\frac{2}{5}}$
$∴ cos 2\theta =1 - 2 sin^2 \theta $
$⇒ cos 2 \theta = 1 - \frac{4}{5}=\frac{1}{5}⇒ \theta = \frac{1}{2} cos^{-1}(\frac{1}{5})$
So, statement-2 is true.