Magnitude of vectors

Solution

Option (a) is correct.

Explanation:

Substituting \( |\vec{a}| = 4 \): \[ |\vec{a} \times \vec{b}|^2 + (|\vec{a}||\vec{b}|\cos\theta)^2 = 144 \] Let \( |\vec{b}| = b \): \[ |\vec{a} \times \vec{b}|^2 + (4b\cos\theta)^2 = 144 \] Using the identity: \[ |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2\theta = 16b^2 \sin^2\theta \] The equation becomes: \[ 16b^2 \sin^2\theta + 16b^2 \cos^2\theta = 144 \] Factoring out \( 16b^2 \): \[ 16b^2 (\sin^2\theta + \cos^2\theta) = 144 \] \[ 16b^2 = 144 \] \[ b^2 = 9 \implies |\vec{b}| = 3 \]