Question

A plane $\pi_1$ contains the vectors $\vec{i}+\vec{j}$ and $\vec{i}+2\vec{j}$. Another plane $\pi_2$ contains the vectors $2\vec{i}-\vec{j}$ and $3\vec{i}+2\vec{k}$. $\vec{a}$ is a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If the angle $\theta$ between $\vec{a}$ and $\vec{i}-2\vec{j}+2\vec{k}$ is acute, then $\theta=$

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\cos^{-1}(\frac{4}{3\sqrt{5}})$
• D. $\cos^{-1}(\frac{2}{\sqrt{5}})$

Hint:

The direction vector of the line of intersection of two planes is always perpendicular to both of their normal vectors. Therefore, it can be found by taking the cross product of the normal vectors: $\vec{d} = \vec{n_1} \times \vec{n_2}$.

Answer 1

The Correct Option is C

Step 1: Find the normal vector to the plane $\pi_1$.
A normal vector $\vec{n_1}$ to plane $\pi_1$ is perpendicular to both vectors lying in it. We can find it using the cross product.
Let $\vec{u} = \vec{i}+\vec{j}$ and $\vec{v} = \vec{i}+2\vec{j}$.
$\vec{n_1} = \vec{u} \times \vec{v} = (\vec{i}+\vec{j}) \times (\vec{i}+2\vec{j}) = \vec{i}\times\vec{i} + 2\vec{i}\times\vec{j} + \vec{j}\times\vec{i} + 2\vec{j}\times\vec{j}$.
$= \vec{0} + 2\vec{k} - \vec{k} + \vec{0} = \vec{k}$.

Step 2: Find the normal vector to the plane $\pi_2$.
A normal vector $\vec{n_2}$ to plane $\pi_2$ is perpendicular to both vectors lying in it.
Let $\vec{p} = 2\vec{i}-\vec{j}$ and $\vec{q} = 3\vec{i}+2\vec{k}$.
$\vec{n_2} = \vec{p} \times \vec{q} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}
2 & -1 & 0
3 & 0 & 2 \end{vmatrix} = \vec{i}(-2) - \vec{j}(4) + \vec{k}(3) = -2\vec{i}-4\vec{j}+3\vec{k}$.

Step 3: Find the direction vector of the line of intersection.
The line of intersection of two planes is perpendicular to both of their normal vectors.
Therefore, a direction vector $\vec{a}$ for this line is parallel to the cross product of the normal vectors, $\vec{n_1} \times \vec{n_2}$.
$\vec{a}$ is parallel to $\vec{k} \times (-2\vec{i}-4\vec{j}+3\vec{k}) = -2(\vec{k}\times\vec{i}) - 4(\vec{k}\times\vec{j}) + 3(\vec{k}\times\vec{k})$.
$= -2(\vec{j}) - 4(-\vec{i}) + \vec{0} = 4\vec{i}-2\vec{j}$.
We can take $\vec{a} = 4\vec{i}-2\vec{j}$ or any scalar multiple, like $2\vec{i}-\vec{j}$. Let's use $\vec{a} = 4\vec{i}-2\vec{j}$.

Step 4: Calculate the angle $\theta$ between $\vec{a$ and the given vector.}
Let $\vec{b} = \vec{i}-2\vec{j}+2\vec{k}$. The angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (4)(1) + (-2)(-2) + (0)(2) = 4 + 4 = 8$.
$|\vec{a}| = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.
$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.
$\cos\theta = \frac{8}{(2\sqrt{5})(3)} = \frac{8}{6\sqrt{5}} = \frac{4}{3\sqrt{5}}$.

Step 5: Ensure the angle is acute.
The value of $\cos\theta = \frac{4}{3\sqrt{5}}$ is positive. Since $\cos\theta$ is positive, the angle $\theta$ must be acute (in the first quadrant). If the dot product had been negative, we would have taken the vector $-\vec{a}$ to ensure an acute angle, but it's not necessary here.
Thus, $\theta = \cos^{-1}(\frac{4}{3\sqrt{5}})$. This matches option (C).