Let \(\vec a =\hat{\imath} + 2\hat{\jmath} + 3\hat{k}\) and \(\vec b=-2\hat{\imath}+ 3\hat{\jmath} - 4\hat{k}\), then which of the following statements are correct?
(A) \(|\vec a| = \sqrt{14}\)
(B) \(|\vec b| = 29\)
(C) \(\vec a.\vec b= 8\)
(D) Angle between \(\vec{a}\) and \(\vec{b}\) is \(\cos^{-1} \left( \displaystyle\frac{-8}{\sqrt{406}} \right)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (D) only
(A) \(|\vec a| = \sqrt{14}\) — Correct. \(|\vec a| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}\).
(B) \(|\vec b| = 29\) — Incorrect. \(|\vec b| = \sqrt{(-2)^2 + 3^2 + (-4)^2} = \sqrt{29}\); 29 is \(|\vec b|^2\).
(C) \(\vec a.\vec b= 8\) — Incorrect. The dot product is \(\vec a \cdot \vec b = -2 + 6 - 12 = -8\).
(D) Angle between \(\vec{a}\) and \(\vec{b}\) is \(\cos^{-1} \left( \frac{-8}{\sqrt{406}} \right)\) — Correct. \(\cos\theta = \frac{-8}{\sqrt{14}\,\sqrt{29}} = \frac{-8}{\sqrt{406}}\).