JEE PYQ: Vector Algebra - Question ID 606309ab4954 (JEE Main 2021)

In vector algebra, the key concepts used here are:

• Cross Product: The cross product of two vectors $\vec{u}$ and $\vec{v}$ yields a vector $\vec{u} \times \vec{v}$ that is perpendicular (normal) to the plane containing $\vec{u}$ and $\vec{v}$. Its magnitude is $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta$, where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$.
• Dot Product: The dot product of two vectors $\vec{u}$ and $\vec{v}$ is $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$. It is used to find the angle between two vectors.
• Unit Vector: A unit vector in the direction of $\vec{v}$ is $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$.

The problem requires finding the angle between two normal vectors, each perpendicular to a different plane defined by pairs of the given vectors.

Step 1: Identify the given vectors

The initial velocity vectors of particles P, Q, and R are: $\overrightarrow{A} = \widehat{i} + \widehat{j}, \quad \overrightarrow{B} = \widehat{j} + \widehat{k}, \quad \overrightarrow{C} = -\widehat{i} + \widehat{j}.$

Step 2: Find the direction of motion of particle P after collision

Particle P moves normal to the plane containing $\overrightarrow{A}$ and $\overrightarrow{B}$. The normal vector to this plane is given by the cross product $\overrightarrow{A} \times \overrightarrow{B}$.

Compute $\overrightarrow{A} \times \overrightarrow{B}$: $$\overrightarrow{A} \times \overrightarrow{B} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{vmatrix} = \widehat{i}(1 \cdot 1 - 0 \cdot 1) - \widehat{j}(1 \cdot 1 - 0 \cdot 0) + \widehat{k}(1 \cdot 1 - 1 \cdot 0) = \widehat{i} - \widehat{j} + \widehat{k}.$$

Thus, the direction of motion of P is along $\overrightarrow{n_P} = \widehat{i} - \widehat{j} + \widehat{k}$.

Step 3: Find the direction of motion of particle Q after collision

Particle Q moves normal to the plane containing $\overrightarrow{A}$ and $\overrightarrow{C}$. The normal vector to this plane is given by the cross product $\overrightarrow{A} \times \overrightarrow{C}$.

Compute $\overrightarrow{A} \times \overrightarrow{C}$: $$\overrightarrow{A} \times \overrightarrow{C} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 1 & 0 \\ -1 & 1 & 0 \\ \end{vmatrix} = \widehat{i}(1 \cdot 0 - 0 \cdot 1) - \widehat{j}(1 \cdot 0 - 0 \cdot (-1)) + \widehat{k}(1 \cdot 1 - 1 \cdot (-1)) = 0\widehat{i} - 0\widehat{j} + 2\widehat{k} = 2\widehat{k}.$$

Thus, the direction of motion of Q is along $\overrightarrow{n_Q} = \widehat{k}$ (the scalar multiple does not affect direction).

Step 4: Compute the angle between $\overrightarrow{n_P}$ and $\overrightarrow{n_Q}$

The angle $\theta$ between $\overrightarrow{n_P}$ and $\overrightarrow{n_Q}$ is given by: $\cos\theta = \frac{\overrightarrow{n_P} \cdot \overrightarrow{n_Q}}{|\overrightarrow{n_P}||\overrightarrow{n_Q}|}.$

Compute the dot product: $\overrightarrow{n_P} \cdot \overrightarrow{n_Q} = (\widehat{i} - \widehat{j} + \widehat{k}) \cdot \widehat{k} = 0 - 0 + 1 = 1.$

Compute the magnitudes: $|\overrightarrow{n_P}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}, \quad |\overrightarrow{n_Q}| = \sqrt{0^2 + 0^2 + 1^2} = 1.$

Thus: $\cos\theta = \frac{1}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}.$

Step 5: Relate to the given angle expression

The problem states that the angle between the directions of P and Q is $\cos^{-1}\left(\frac{1}{\sqrt{x}}\right)$. From Step 4, we have: $\cos\theta = \frac{1}{\sqrt{3}}.$

Comparing the two expressions: $\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{3}} \implies \sqrt{x} = \sqrt{3} \implies x = 3.$

1. Incorrect Cross Product Calculation: Students often make sign errors or forget the order of operations in the determinant expansion. Always double-check the cross product using the right-hand rule or re-derive it.
2. Ignoring Unit Vectors: The problem asks for the angle between directions, not the vectors themselves. While magnitudes are used in the dot product formula, the direction is what matters. Scalars (like the 2 in $\overrightarrow{n_Q}$) do not affect the angle.
3. Misinterpreting the Normal Vector: Some students assume the normal vector must be in a specific direction (e.g., "up" or "down"). The cross product can yield either of the two possible normals, but the angle between the two normals will be the same regardless of direction.
4. Overcomplicating the Problem: The question is purely about vector algebra. Avoid introducing unnecessary physics (e.g., momentum conservation) unless explicitly asked.

Final Answer: The value of $x$ is $\boxed{3}$.