The angle between the vectors imath-2jmath+k and 3imath+2jmath+k is:

The correct answer is Option (4) → \(\displaystyle\frac{\pi}{2}\)

Let \(\vec{a} = \hat{\imath}-2\hat{\jmath}+\hat{k}\) and \(\vec{b} = 3\hat{\imath}+2\hat{\jmath}+\hat{k}\). The angle comes from the dot product:

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

    \(\vec{a}\cdot\vec{b} = (1)(3)+(-2)(2)+(1)(1)\)

    \(= 3-4+1 = 0\)

Both vectors are non-zero, so \(\cos\theta = 0\) and

    \(\displaystyle \theta = \frac{\pi}{2}\)

The vectors are perpendicular; the magnitudes are not needed once the dot product is 0.