Key Concepts and Formulas

1. Normal Vector to a Plane: If a plane contains two non-collinear vectors $\vec{u}$ and $\vec{v}$, then a vector normal (perpendicular) to the plane is given by their cross product: $\vec{n} = \vec{u} \times \vec{v}$. This vector is orthogonal to every vector in the plane.
2. Direction Vector of the Line of Intersection of Two Planes: The line formed by the intersection of two planes is perpendicular to the normal vector of the first plane and also to the normal vector of the second plane. Therefore, its direction vector is parallel to the cross product of the two normal vectors: $\vec{d} = \vec{n_1} \times \vec{n_2}$.
3. Angle Between Two Vectors: The angle $\theta$ between two non-zero vectors $\vec{A}$ and $\vec{B}$ is determined by the dot product formula: $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$. If an obtuse angle is required and the calculated $\cos \theta$ is positive (indicating an acute angle $\theta_{acute}$), the obtuse angle is $\pi - \theta_{acute}$.

Step-by-Step Solution

Our objective is to find the obtuse angle between a vector $\vec{a}$ (which is parallel to the line of intersection of two given planes) and the vector $\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$.

Step 1: Determine the Normal Vector to the First Plane ($\vec{n_1}$)

Why this step? To find the direction vector of the line of intersection, we first need the normal vectors of the two planes. A normal vector to a plane can be found by taking the cross product of any two non-collinear vectors lying within that plane. The first plane is defined by the vectors $\vec{u_1} = \hat{i}$ and $\vec{v_1} = \hat{i}+\hat{j}$. Let $\vec{n_1}$ be the normal vector to this plane: $\vec{n_1} = \vec{u_1} \times \vec{v_1} = \hat{i} \times (\hat{i}+\hat{j})$ Using the properties of cross products ($\hat{i} \times \hat{i} = \vec{0}$ and $\hat{i} \times \hat{j} = \hat{k}$): $\vec{n_1} = (\hat{i} \times \hat{i}) + (\hat{i} \times \hat{j}) = \vec{0} + \hat{k} = \hat{k}$ Thus, the normal vector to the first plane is $\vec{n_1} = \hat{k}$.

Step 2: Determine the Normal Vector to the Second Plane ($\vec{n_2}$)

Why this step? Similar to Step 1, we need the normal vector for the second plane to ultimately determine the direction of the line of intersection. The second plane is defined by the vectors $\vec{u_2} = \hat{i}-\hat{j}$ and $\vec{v_2} = \hat{i}+\hat{k}$. Let $\vec{n_2}$ be the normal vector to this plane: $\vec{n_2} = \vec{u_2} \times \vec{v_2} = (\hat{i}-\hat{j}) \times (\hat{i}+\hat{k})$ We calculate this using the determinant form: $\vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 0 & 1 \end{vmatrix}$ $\vec{n_2} = \hat{i}((-1)(1) - (0)(0)) - \hat{j}((1)(1) - (0)(1)) + \hat{k}((1)(0) - (-1)(1))$ $\vec{n_2} = \hat{i}(-1) - \hat{j}(1) + \hat{k}(1) = -\hat{i} - \hat{j} + \hat{k}$ Thus, the normal vector to the second plane is $\vec{n_2} = -\hat{i} - \hat{j} + \hat{k}$.

Step 3: Determine the Direction Vector of the Line of Intersection ($\vec{a}$)

Why this step? The vector $\vec{a}$ is parallel to the line of intersection. This line is perpendicular to both normal vectors ($\vec{n_1}$ and $\vec{n_2}$). Therefore, its direction vector is parallel to the cross product of these two normal vectors. Let $\vec{a}$ be the direction vector of the line of intersection: $\vec{a} = \vec{n_1} \times \vec{n_2}$ Substitute the normal vectors found in Step 1 and Step 2: $\vec{a} = \hat{k} \times (-\hat{i} - \hat{j} + \hat{k})$ Expand the cross product using distributive property and cyclic order ($\hat{k} \times \hat{i} = \hat{j}$, $\hat{k} \times \hat{j} = -\hat{i}$, $\hat{k} \times \hat{k} = \vec{0}$): $\vec{a} = \hat{k} \times (-\hat{i}) + \hat{k} \times (-\hat{j}) + \hat{k} \times \hat{k}$ $\vec{a} = -(\hat{k} \times \hat{i}) - (\hat{k} \times \hat{j}) + \vec{0}$ $\vec{a} = -\hat{j} - (-\hat{i}) = -\hat{j} + \hat{i}$ So, the vector $\vec{a} = \hat{i} - \hat{j}$.

Step 4: Calculate the Angle Between $\vec{a}$ and $\vec{b}$

Why this step? We use the dot product formula to find the cosine of the angle between $\vec{a}$ and $\vec{b}$, which is the core part of finding the angle. We have $\vec{a} = \hat{i} - \hat{j}$ and the given vector $\vec{b} = \hat{i}-2 \hat{j}+2 \hat{k}$. Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$. The formula is $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.

First, calculate the dot product $\vec{a} \cdot \vec{b}$: $\vec{a} \cdot \vec{b} = (\hat{i} - \hat{j}) \cdot (\hat{i}-2 \hat{j}+2 \hat{k}) = (1)(1) + (-1)(-2) + (0)(2) = 1 + 2 + 0 = 3$

Next, calculate the magnitudes of $\vec{a}$ and $\vec{b}$: $|\vec{a}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1+1} = \sqrt{2}$ $|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$

Now, substitute these values into the cosine formula: $\cos \theta = \frac{3}{(\sqrt{2})(3)} = \frac{1}{\sqrt{2}}$

Step 5: Determine the Obtuse Angle

Why this step? The problem specifically asks for the obtuse angle. The value $\cos \theta = \frac{1}{\sqrt{2}}$ corresponds to an acute angle. From $\cos \theta = \frac{1}{\sqrt{2}}$, the acute angle is $\theta_{acute} = \frac{\pi}{4}$. Since the problem asks for the obtuse angle, we use the relationship: $\theta_{obtuse} = \pi - \theta_{acute}$ $\theta_{obtuse} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

Common Mistakes & Tips

• Cross Product Errors: Be meticulous with the signs and order when calculating cross products. Remember the cyclic order for unit vectors: $\hat{i} \to \hat{j} \to \hat{k} \to \hat{i}$. Going against the cycle introduces a negative sign (e.g., $\hat{j} \times \hat{i} = -\hat{k}$).
• Acute vs. Obtuse Angle: Always double-check if the question asks for an acute or obtuse angle. If $\cos \theta > 0$, the direct result from $\arccos(\cos \theta)$ is acute. If an obtuse angle is needed, subtract the acute angle from $\pi$. If $\cos \theta < 0$, the direct result is already obtuse.
• Magnitude Calculations: Ensure accurate calculation of vector magnitudes to avoid errors in the final cosine value.

Summary

This problem required a sequential application of vector operations. First, we found the normal vectors to the two planes using the cross product of the vectors defining each plane. Then, we found the direction vector $\vec{a}$ of their line of intersection by taking the cross product of these two normal vectors. Finally, we used the dot product formula to find the cosine of the angle between $\vec{a}$ and $\vec{b}$, and then adjusted the acute angle to find the required obtuse angle. The calculated obtuse angle is $\frac{3\pi}{4}$.

The final answer is $\boxed{\text{A}}$ which corresponds to option (A).