Let \(\theta\) be the angle between the planes \(P_{1}=\vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9\) and \(P_{2}=\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15\). Let L be the line that meets \(\mathrm{P}_{2}\) at the point ( \(4,-2,5\) ) and makes an angle \(\theta\) with the normal of \(\mathrm{P}_{2}\). If \(\alpha\) is the angle between L and \(\mathrm{P}_{2}\) then \(\left(\tan ^{2} \theta\right)\left(\cot ^{2} \alpha\right)\) is equal to

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Three Dimensional Geometry

Let

θ

be the angle between the planes

P_{1} = r \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 9

and

P_{2} = r \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 15

. Let L be the line that meets

P_{2}

at the point (

4, - 2, 5

) and makes an angle

θ

with the normal of

P_{2}

. If

α

is the angle between L and

P_{2}

then

(tan^{2} θ) (cot^{2} α)

is equal to

Answer: 9

The plane normals are

n_{1} = \hat{i} + \hat{j} + 2 \hat{k}

and

n_{2} = 2 \hat{i} - \hat{j} + \hat{k}

. Their angle satisfies

cos θ = \frac{n _{1} \cdot n _{2}}{∣ n _{1} ∣∣ n _{2} ∣} = \frac{3}{6} = \frac{1}{2}

, so

θ = \frac{π}{3}

. The line

L

makes this same angle

θ

with the normal to

P_{2}

, and the angle a line makes with a plane is the complement of the angle it makes with that normal, so

α = \frac{π}{2} - θ = \frac{π}{6}

. Therefore

tan^{2} θ cot^{2} α = (3)^{2} (3)^{2} = 9

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer