Let \(\vec{a}=\hat{\imath}+2\hat{\jmath}-\hat{k}\) and \(\vec{b}=3\hat{\imath}-\hat{\jmath}+2\hat{k}\). Which of the following statements are correct?
(A) \(\vec{a}+\vec{b} = 4\hat{\imath}+3\hat{\jmath}+\hat{k}\)
(B) \(\vec{a}-\vec{b}\) is collinear with \(-4\hat{\imath}+6\hat{\jmath}-6\hat{k}\)
(C) \(|\vec{a}| = |\vec{b}|\)
(D) \(|\vec{a}+\vec{b}| = 3\sqrt{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (D) only
Adding and subtracting component by component:
\(\vec{a}+\vec{b} = 4\hat{\imath}+\hat{\jmath}+\hat{k}\)
\(\vec{a}-\vec{b} = -2\hat{\imath}+3\hat{\jmath}-3\hat{k}\)
The magnitudes are
\(|\vec{a}| = \sqrt{1+4+1} = \sqrt{6}\)
\(|\vec{b}| = \sqrt{9+1+4} = \sqrt{14}\)
\(|\vec{a}+\vec{b}| = \sqrt{16+1+1} = \sqrt{18} = 3\sqrt{2}\)
So \(\vec{a}+\vec{b} = 4\hat{\imath}+\hat{\jmath}+\hat{k}\), \(\vec{a}-\vec{b} = -2\hat{\imath}+3\hat{\jmath}-3\hat{k}\), \(|\vec{a}| = \sqrt{6}\), \(|\vec{b}| = \sqrt{14}\) and \(|\vec{a}+\vec{b}| = 3\sqrt{2}\).
Checking the statements
(A) \(\vec{a}+\vec{b} = 4\hat{\imath}+3\hat{\jmath}+\hat{k}\) — Incorrect. The \(\hat{\jmath}\)-component is \(2-1 = 1\), not 3.
(B) \(\vec{a}-\vec{b}\) is collinear with \(-4\hat{\imath}+6\hat{\jmath}-6\hat{k}\) — Correct. That vector is \(2(\vec{a}-\vec{b})\).
(C) \(|\vec{a}| = |\vec{b}|\) — Incorrect. \(|\vec{a}| = \sqrt{6}\) but \(|\vec{b}| = \sqrt{14}\).
(D) \(|\vec{a}+\vec{b}| = 3\sqrt{2}\) — Correct. \(\sqrt{16+1+1} = \sqrt{18} = 3\sqrt{2}\).