The problem asks for the value of x, given the angle between the directions of motion of particles P and Q.
Finding Particle Motion Directions
Particle P moves normal to the plane containing vectors $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + \hat{k}$. This direction is found using the cross product $\vec{N_P} = \vec{A} \times \vec{B}$.
$ \vec{N_P} = \vec{A} \times \vec{B} = (\hat{i} + \hat{j}) \times (\hat{j} + \hat{k}) $ $ \vec{N_P} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 0 \cdot 1) - \hat{j}(1 \cdot 1 - 0 \cdot 0) + \hat{k}(1 \cdot 1 - 1 \cdot 0) $ $ \vec{N_P} = \hat{i} - \hat{j} + \hat{k} $ The magnitude is $|\vec{N_P}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
Particle Q moves normal to the plane containing vectors $\vec{A} = \hat{i} + \hat{j}$ and $\vec{C} = -\hat{i} + \hat{j}$. This direction is found using the cross product $\vec{N_Q} = \vec{A} \times \vec{C}$.
$ \vec{N_Q} = \vec{A} \times \vec{C} = (\hat{i} + \hat{j}) \times (-\hat{i} + \hat{j}) $ $ \vec{N_Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ -1 & 1 & 0 \end{vmatrix} = \hat{i}(1 \cdot 0 - 0 \cdot 1) - \hat{j}(1 \cdot 0 - 0 \cdot (-1)) + \hat{k}(1 \cdot 1 - 1 \cdot (-1)) $ $ \vec{N_Q} = \hat{i}(0) - \hat{j}(0) + \hat{k}(1 + 1) = 2\hat{k} $ The magnitude is $|\vec{N_Q}| = \sqrt{0^2 + 0^2 + 2^2} = 2$.
Calculating Angle Between Motion Directions
The angle $\theta$ between the directions of motion $\vec{N_P}$ and $\vec{N_Q}$ is determined using the dot product formula:
$ \cos \theta = \frac{\vec{N_P} \cdot \vec{N_Q}}{|\vec{N_P}| |\vec{N_Q}|} $
First, calculate the dot product:
$ \vec{N_P} \cdot \vec{N_Q} = (\hat{i} - \hat{j} + \hat{k}) \cdot (2\hat{k}) = (1)(0) + (-1)(0) + (1)(2) = 2 $
Now, substitute the dot product and magnitudes into the cosine formula:
$ \cos \theta = \frac{2}{\sqrt{3} \times 2} = \frac{1}{\sqrt{3}} $
Solving for x
The problem states that the angle between the directions is $\cos^{-1}\left(\frac{1}{\sqrt{x}}\right)$, which means $\cos \theta = \frac{1}{\sqrt{x}}$.
By equating the calculated value of $\cos \theta$ with the given expression:
$ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{x}} $
Solving for $x$ gives:
$ \sqrt{x} = \sqrt{3} \implies x = 3 $
The value of $x$ is 3.
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Important Questions from General Physics
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