Match List-I with List-II Consider two vectors a =imath+2jmath-k and…

The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

The angle a vector makes with an axis comes from its direction cosines: each component divided by the magnitude.

    \(|\vec{a}| = \sqrt{1 + 4 + 1} = \sqrt{6}\)

    \(\vec{b} = -3\vec{a},\quad |\vec{b}| = 3\sqrt{6}\)

(A) Angle between \(\vec{a}\) and \(\vec{b}\)

This matches (III) \(\pi\). Since \(\vec{b}\) is a negative multiple of \(\vec{a}\), the two vectors point in opposite directions:

    \(\displaystyle \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\)

    \(\displaystyle = \frac{-3 - 12 - 3}{\sqrt{6}\cdot 3\sqrt{6}} = \frac{-18}{18} = -1\)

    \(\theta = \pi\)

(B) Angle between \(\vec{a}\) and x-axis

This matches (I). The x-component of \(\vec{a}\) is 1:

    \(\displaystyle \cos\alpha = \frac{1}{\sqrt{6}},\quad \alpha = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right)\)

(C) Angle between \(\vec{b}\) and x-axis

This matches (IV). The x-component of \(\vec{b}\) is −3:

    \(\displaystyle \cos\alpha' = \frac{-3}{3\sqrt{6}} = -\frac{1}{\sqrt{6}},\quad \alpha' = \cos^{-1}\left(-\frac{1}{\sqrt{6}}\right)\)

(D) Angle between \(\vec{a}\) and y-axis

This matches (II). The y-component of \(\vec{a}\) is 2:

    \(\displaystyle \cos\beta = \frac{2}{\sqrt{6}},\quad \beta = \cos^{-1}\left(\frac{2}{\sqrt{6}}\right)\)

So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).