Question

Let \[ \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}, \quad \text{where } a_1, a_2, a_3 \text{ and } |\vec{a}| \text{ are rational numbers.} \] If \(\vec{a}\) makes an angle of \(45^\circ\) with \[ \vec{b} = \sqrt{2} \hat{i} + 3 \sqrt{2} \hat{j} + 4 \hat{k}, \] then \(\vec{a}\) lies in which plane?

• A. XY-plane
• B. YZ-plane
• C. XZ-plane
• D. Along the bisector of the angle between \(\hat{k}\) and \(\vec{b}\)

Hint:

For vectors with rational magnitude and dot product, any irrational coefficient in one vector forces the corresponding component in the other vector to be zero.

Answer 1

The Correct Option is A

Step 1: Recall dot product formula.
\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]

Step 2: Substitute \(\theta = 45^\circ\).
\[ \cos 45^\circ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{\sqrt{2}} \]

Step 3: Express \(\vec{a} \cdot \vec{b}\).
\[ \vec{a} \cdot \vec{b} = a_1 \cdot \sqrt{2} + a_2 \cdot 3\sqrt{2} + a_3 \cdot 4 \] Since \(|\vec{a}|\) is rational, \(\vec{a} \cdot \vec{b}\) must also be rational.

Step 4: Rationality condition.
For \(\vec{a} \cdot \vec{b}\) to be rational, \(a_3\) must be \(0\), because \(\sqrt{2}\) is irrational.

Step 5: Conclude plane.
If \(a_3 = 0\), \(\vec{a}\) has only \(x\) and \(y\) components, so it lies in the XY-plane.

Step 6: Final conclusion.
\[ \boxed{\text{XY-plane}} \]