Question

Let $\theta$ be the angle between the planes $P_1: \vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9$ and $P_2: \hat{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15$ Let $L$ be the line that meets $P_2$ at the point $(4,-2,5)$ and makes an angle $\theta$ with the normal of $P_4$ If $\alpha$ is the angle between $L$ and $P_2$, then $\left(\tan ^2 \theta\right)\left(\cot ^2 \alpha\right)$ is equal to

Hint:

When calculating angles between planes and lines, use the dot product to find the cosine of the angle between their normals. The angle between the line and the plane can then be calculated using the complementary angle relation.

Answer 1

Correct Answer: 9

The correct answer is 9.

$\cos \theta =\frac{(\hat{i}+\hat{j}+2\hat{k})\cdot (2\hat{i}-\hat{j}+\hat{k})}{6}=\frac{2-1+2}{6}=\frac{1}{2}$
$\theta =\pi /3$
$\alpha =\pi /6$
$\left({\tan }^2\theta \right)\left({\cot }^2\alpha \right)$
$(3)(3)=9$

Answer 2

Step 1: Compute the Angle Between Two Vectors

Given two vectors:

\[ \mathbf{A} = \hat{i} + \hat{j} + 2\hat{k}, \quad \mathbf{B} = 2\hat{i} - \hat{j} + \hat{k} \]

The dot product formula is:

\[ \mathbf{A} \cdot \mathbf{B} = (1 \times 2) + (1 \times -1) + (2 \times 1) \] \[ = 2 - 1 + 2 = 3. \]

The magnitudes of the vectors are:

\[ |\mathbf{A}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}, \quad |\mathbf{B}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}. \]

Thus, the cosine of the angle between them is:

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{3}{6} = \frac{1}{2}. \]

So, we get:

\[ \theta = \frac{\pi}{3}. \]

Step 2: Compute Another Angle

Given that:

\[ \alpha = \frac{\pi}{6}. \]

Step 3: Compute the Expression

Now, we calculate:

\[ (\tan^2 \theta)(\cot^2 \alpha). \]

Using the values:

\[ \tan^2 \theta = \tan^2 \frac{\pi}{3} = (3)^2 = 9. \] \[ \cot^2 \alpha = \cot^2 \frac{\pi}{6} = (3)^2 = 9. \]

Thus,

\[ (\tan^2 \theta)(\cot^2 \alpha) = 9 \times 9 = 81. \]

Final Answer:

The result of the given expression is:

\[ 81. \]