The projection of the vector \(\vec{a} =\hat{\imath}+2\hat{\jmath}-3\hat{k}\) on the vector \(2\hat{\imath} +6\hat{\jmath}+3\hat{k}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{5}{7}\)
Let \(\vec{b} = 2\hat{\imath} + 6\hat{\jmath} + 3\hat{k}\). The projection of a on b is
\(\displaystyle \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\)
\(\vec{a}\cdot\vec{b} = (1)(2) + (2)(6) + (-3)(3)\)
\(= 2 + 12 - 9 = 5\)
\(\displaystyle |\vec{b}| = \sqrt{4 + 36 + 9} = \sqrt{49} = 7\)
\(\displaystyle \text{Projection} = \frac{5}{7}\)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{5}{\sqrt{14}}\) — Incorrect. This divides by \(|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}\), so it is the projection of \(\vec{b}\) on \(\vec{a}\). The divisor is the length of the vector projected on.
Option (3) 0 — Incorrect. The projection is 0 only for perpendicular vectors, but \(\vec{a}\cdot\vec{b} = 5 \ne 0\).
Option (4) \(\displaystyle\frac{5}{14}\) — Incorrect. The divisor is \(|\vec{b}| = 7\), not 14.