A vector of magnitude 14 has direction ratios 2, 3, −6 and makes an acute angle with the positive direction of the x-axis. The vector is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4\hat{\imath}+6\hat{\jmath}-12\hat{k}\)
Since the direction ratios are 2, 3, −6, the vector is a multiple of \(2\hat{\imath}+3\hat{\jmath}-6\hat{k}\):
\(\vec{r} = k(2\hat{\imath}+3\hat{\jmath}-6\hat{k})\)
\(|\vec{r}| = |k|\sqrt{4+9+36} = 7|k|\)
\(7|k| = 14 \;\Rightarrow\; k = \pm 2\)
An acute angle with the x-axis needs a positive x-component, so \(k = 2\):
\(\vec{r} = 4\hat{\imath}+6\hat{\jmath}-12\hat{k}\)
Why the other options are incorrect
Option (1) \(-4\hat{\imath}-6\hat{\jmath}+12\hat{k}\) — Incorrect. Its length is 14, but its x-component is negative, so the angle with the x-axis is obtuse.
Option (2) \(\frac{2}{7}\hat{\imath}+\frac{3}{7}\hat{\jmath}-\frac{6}{7}\hat{k}\) — Incorrect. This is the unit vector in that direction; its length is 1, not 14.
Option (4) \(28\hat{\imath}+42\hat{\jmath}-84\hat{k}\) — Incorrect. This is 14 times the direction ratios; its length is \(14 \times 7 = 98\).