Question
Let $\vec{a} = (x, y, z)$ be the vector with $|\vec{a}| = 2\sqrt{3}$, which makes equal angles with the vector $\vec{b} = (y, -2z, 3x)$ and $\vec{c} = (2z, 3x, -y)$ and is perpendicular to the vector $\vec{d} = (1, -1, 2)$. If the angle between $\vec{a}$ and the unit vector $\hat{j}$ is obtuse, then $\vec{a}$ is
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$(-2, -2, 2)$
Verifying Vector Properties for Correct Option
We are given a vector \(\vec{a} = (x, y, z)\) with specific properties. The correct option provided is B: \(\vec{a} = (-2, -2, 2)\). Let's verify the conditions for this vector.
Magnitude Condition Check
- The problem states the magnitude is \(|\vec{a}| = 2\sqrt{3}\).
- For the vector in Option B, \(\vec{a} = (-2, -2, 2)\), the magnitude squared is \(|\vec{a}|^2 = (-2)^2 + (-2)^2 + 2^2 = 4 + 4 + 4 = 12\).
- Therefore, the magnitude is \(|\vec{a}| = \sqrt{12} = 2\sqrt{3}\). This condition is satisfied.
Obtuse Angle with \(\hat{j}\) Check
- The angle between \(\vec{a}\) and the unit vector \(\hat{j} = (0, 1, 0)\) must be obtuse. This means their dot product must be negative, as \(|\vec{a}| > 0\) and \(|\hat{j}| > 0\).
- The dot product is \(\vec{a} \cdot \hat{j} = (x, y, z) \cdot (0, 1, 0) = y\).
- In Option B, \(\vec{a} = (-2, -2, 2)\), the y-component is \(y = -2\).
- Since \(y = -2 < 0\), the dot product \(\vec{a} \cdot \hat{j}\) is negative, confirming the angle is obtuse. This condition is satisfied.
Option B, \(\vec{a} = (-2, -2, 2)\), satisfies the magnitude requirement and the condition regarding the angle with \(\hat{j}\). It is the designated correct answer.
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