Let \(\vec{a}=\hat{\imath}-4\hat{\jmath}+8\hat{k}\). Which of the following statements are correct?
(A) \(|\vec{a}| = 9\)
(B) The direction cosines of \(\vec{a}\) are \(\displaystyle\frac{1}{9}, -\frac{4}{9}, \frac{8}{9}\)
(C) \(\vec{a}\) is collinear with the vector \(-3\hat{\imath}+12\hat{\jmath}-24\hat{k}\)
(D) The direction cosines of \(3\vec{a}\) are \(\displaystyle\frac{1}{3}, -\frac{4}{3}, \frac{8}{3}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (C) only
The magnitude of \(\vec{a}\) is
\(|\vec{a}| = \sqrt{1+16+64} = \sqrt{81} = 9\)
The direction cosines are the components divided by 9:
\( \displaystyle \frac{1}{9}, -\frac{4}{9}, \frac{8}{9} \)
The vector in (C) is a scalar multiple of \(\vec{a}\):
\(-3\hat{\imath}+12\hat{\jmath}-24\hat{k} = -3\vec{a}\)
The vector \(3\vec{a}\) has the same direction as \(\vec{a}\), so it has the same direction cosines.
So \(|\vec{a}| = 9\), the direction cosines of both \(\vec{a}\) and \(3\vec{a}\) are \(\frac{1}{9}, -\frac{4}{9}, \frac{8}{9}\), and \(-3\hat{\imath}+12\hat{\jmath}-24\hat{k} = -3\vec{a}\).
Checking the statements
(A) \(|\vec{a}| = 9\) — Correct. \(\sqrt{1+16+64} = 9\).
(B) The direction cosines of \(\vec{a}\) are \(\frac{1}{9}, -\frac{4}{9}, \frac{8}{9}\) — Correct. Each component is divided by \(|\vec{a}| = 9\).
(C) \(\vec{a}\) is collinear with \(-3\hat{\imath}+12\hat{\jmath}-24\hat{k}\) — Correct. That vector is \(-3\vec{a}\).
(D) The direction cosines of \(3\vec{a}\) are \(\frac{1}{3}, -\frac{4}{3}, \frac{8}{3}\) — Incorrect. They are \(\frac{1}{9}, -\frac{4}{9}, \frac{8}{9}\); direction cosines can never exceed 1 in size.