Q78 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics

The acute angle $\theta$ between the $xy$-plane and the plane passing through the point $(1, 2, 4)$ and parallel to the vectors with direction ratios $3, 2, -1$ and $1, -2, -2$ is...

  • Correct answer
    A
    $\cos^{-1}\left(\dfrac{8}{5\sqrt{5}}\right)$
  • B
    $\cos^{-1}\left(\dfrac{5}{8\sqrt{5}}\right)$
  • C
    $\sin^{-1}\left(\dfrac{8}{5\sqrt{5}}\right)$
  • D
    $\dfrac{\pi}{4}$

Solution

Let the vectors parallel to the plane be $\vec{v}_1 = 3\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{v}_2 = \hat{i} - 2\hat{j} - 2\hat{k}$.

The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{v}_1$ and $\vec{v}_2$:

$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 1 & -2 & -2 \end{vmatrix}$

$\vec{n} = \hat{i}(-4 - 2) - \hat{j}(-6 - (-1)) + \hat{k}(-6 - 2) = -6\hat{i} + 5\hat{j} - 8\hat{k}$

The normal vector to the $xy$-plane is $\hat{k}$.

The acute angle $\theta$ between the two planes is the acute angle between their normal vectors, given by:

$\cos \theta = \dfrac{|\vec{n} \cdot \hat{k}|}{|\vec{n}| |\hat{k}|}$

$\cos \theta = \dfrac{|(-6\hat{i} + 5\hat{j} - 8\hat{k}) \cdot \hat{k}|}{\sqrt{(-6)^2 + 5^2 + (-8)^2} \sqrt{1^2}}$

$\cos \theta = \dfrac{|-8|}{\sqrt{36 + 25 + 64}} = \dfrac{8}{\sqrt{125}} = \dfrac{8}{5\sqrt{5}}$

$\theta = \cos^{-1}\left(\dfrac{8}{5\sqrt{5}}\right)$

Answer: $\cos^{-1}\left(\dfrac{8}{5\sqrt{5}}\right)$

Hence, the correct answer is option: 'A'.

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