Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
A vector $\vec{a} = a\vec{i} + b\vec{j} + c\vec{k}$ is said to be rational vector. If $a, b, c$ are all rational. If a rational vector with magnitude as positive integer makes an angle $\pi/4$ with vector $\vec{b} = \sqrt{2}\vec{i} + \sqrt{2}\vec{j} + \vec{k}$, then $\vec{a}$:
Surely lies in xy plane
Surely lies in xz plane
Surely lies in yz plane
Nothing can be said
Step-by-Step Solution
Key Concept: Rational coefficient conditions force irrational terms to cancel, constraining the vector to a plane.
Given $\frac{1}{\sqrt{2}} = \frac{\sqrt{2a} + 3\sqrt{2b} + 4c}{6m}$ with $m = |\vec{a}| = 2(3a + 3b) + 4\sqrt{2}c = 6m$, rational values of $a, b, c$ require the coefficient of $\sqrt{2}$ to vanish. This means $c = 0$, so the vector lies in the $xy$-plane.
Correct Answer: 1