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Step-by-Step Solution
Step 1: Identify the given vectors
We have three given vectors:
\overrightarrow{A} = \hat{i} + \hat{j}, \quad
\overrightarrow{B} = \hat{j} + \hat{k}, \quad
\overrightarrow{C} = -\hat{i} + \hat{j}.
Step 2: Find the direction of motion for particle P
Particle P moves normal to the plane containing vectors \overrightarrow{A} and \overrightarrow{B} . The direction normal to a plane formed by two vectors can be found using their cross product. So, for P:
Compute \overrightarrow{A} \times \overrightarrow{B} :
\overrightarrow{A} \times \overrightarrow{B} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 0 \\
0 & 1 & 1
\end{vmatrix}
Using the determinant expansion:
\overrightarrow{A} \times \overrightarrow{B}
= \hat{i}(1 \cdot 1 - 0 \cdot 1)
- \hat{j}(1 \cdot 1 - 0 \cdot 0)
+ \hat{k}(1 \cdot 1 - 1 \cdot 0)
= \hat{i}(1) \;-\; \hat{j}(1) \;+\; \hat{k}(1)
= \hat{i} - \hat{j} + \hat{k}.
Now, the magnitude of \overrightarrow{A} \times \overrightarrow{B} is:
\bigl|\overrightarrow{A} \times \overrightarrow{B}\bigr|
= \sqrt{1^2 + (-1)^2 + 1^2}
= \sqrt{1 + 1 + 1}
= \sqrt{3}.
Hence, the unit vector normal to the plane containing \overrightarrow{A} and \overrightarrow{B} is
\hat{n}_{1}
= \frac{\overrightarrow{A} \times \overrightarrow{B}}{\bigl|\overrightarrow{A} \times \overrightarrow{B}\bigr|}
= \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}.
Step 3: Find the direction of motion for particle Q
Particle Q moves normal to the plane containing vectors \overrightarrow{A} and \overrightarrow{C} . Similarly, we calculate \overrightarrow{A} \times \overrightarrow{C} :
\overrightarrow{A} \times \overrightarrow{C}
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 0 \\
-1 & 1 & 0
\end{vmatrix}.
Expanding this:
\overrightarrow{A} \times \overrightarrow{C}
= \hat{i}(1 \cdot 0 - 0 \cdot 1)
- \hat{j}(1 \cdot 0 - 0 \cdot (-1))
+ \hat{k}(1 \cdot 1 - 1 \cdot (-1))
= \hat{i}(0) \;-\; \hat{j}(0)
+ \hat{k}(1 + 1)
= 2\,\hat{k}.
The magnitude of \overrightarrow{A} \times \overrightarrow{C} is
\bigl|\overrightarrow{A} \times \overrightarrow{C}\bigr|
= \sqrt{2^2}
= 2.
Thus, the unit vector normal to the plane containing \overrightarrow{A} and \overrightarrow{C} is
\hat{n}_{2}
= \frac{\overrightarrow{A} \times \overrightarrow{C}}{\bigl|\overrightarrow{A} \times \overrightarrow{C}\bigr|}
= \frac{2\,\hat{k}}{2}
= \hat{k}.
Step 4: Find the angle between the directions of P and Q
The angle \theta between the two directions \hat{n}_1 and \hat{n}_2 is given by
\cos \theta = \hat{n}_1 \cdot \hat{n}_2.
We have
\hat{n}_1 = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}, \quad
\hat{n}_2 = \hat{k}.
Therefore,
\hat{n}_1 \cdot \hat{n}_2
= \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \cdot \hat{k}
= \frac{1}{\sqrt{3}}\bigl(\hat{i} \cdot \hat{k}
- \hat{j} \cdot \hat{k}
+ \hat{k} \cdot \hat{k}\bigr).
Since \hat{i} \cdot \hat{k} = 0 and \hat{j} \cdot \hat{k} = 0 , while \hat{k} \cdot \hat{k} = 1 , we get
\hat{n}_1 \cdot \hat{n}_2 = \frac{1}{\sqrt{3}}.
Hence,
\cos \theta = \frac{1}{\sqrt{3}}.
Step 5: Relate the angle to \cos^{-1}(1/\sqrt{x})
We are told the angle between the directions of motion of P and Q is
\theta = \cos^{-1}\bigl(\frac{1}{\sqrt{x}}\bigr).
Since we have found
\cos \theta = \frac{1}{\sqrt{3}},
we compare and see:
\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{3}}
\;\;\Longrightarrow\;\;
\sqrt{x} = \sqrt{3}
\;\;\Longrightarrow\;\;
x = 3.
Final Answer
\boxed{3}
