Let \(\vec{a}=\hat{\imath}+2\hat{\jmath}-2\hat{k}\) and \(\vec{b}=2\hat{\imath}+\hat{\jmath}+2\hat{k}\). Which of the following statements are correct?
(A) \(\vec{a}\) and \(\vec{b}\) are perpendicular
(B) \(|\vec{a}\times\vec{b}| = 9\)
(C) The projection of \(\vec{a}\) on \(\vec{b}\) is 1
(D) \(|\vec{a}+\vec{b}| = 3\sqrt{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
The scalar product is
\(\vec{a}\cdot\vec{b} = 2+2-4 = 0\)
The magnitudes are
\(|\vec{a}| = \sqrt{1+4+4} = 3, \quad |\vec{b}| = \sqrt{4+1+4} = 3\)
The vector product is
\( \displaystyle \vec{a}\times\vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & -2 \\ 2 & 1 & 2 \end{vmatrix} \)
\( = \hat{\imath}(4+2)-\hat{\jmath}(2+4)+\hat{k}(1-4)\)
\( = 6\hat{\imath}-6\hat{\jmath}-3\hat{k}\)
\(|\vec{a}\times\vec{b}| = \sqrt{36+36+9} = 9\)
The sum is
\(\vec{a}+\vec{b} = 3\hat{\imath}+3\hat{\jmath}\)
\(|\vec{a}+\vec{b}| = \sqrt{18} = 3\sqrt{2}\)
So \(\vec{a}\cdot\vec{b} = 0\), \(|\vec{a}\times\vec{b}| = 9\), the projection of \(\vec{a}\) on \(\vec{b}\) is 0, and \(|\vec{a}+\vec{b}| = 3\sqrt{2}\).
Checking the statements
(A) \(\vec{a}\) and \(\vec{b}\) are perpendicular — Correct. \(\vec{a}\cdot\vec{b} = 2+2-4 = 0\).
(B) \(|\vec{a}\times\vec{b}| = 9\) — Correct. \(\sqrt{36+36+9} = 9\), the same as \(3\times 3\times\sin 90^{\circ}\).
(C) The projection of \(\vec{a}\) on \(\vec{b}\) is 1 — Incorrect. It is \(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \frac{0}{3} = 0\).
(D) \(|\vec{a}+\vec{b}| = 3\sqrt{2}\) — Correct. \(|3\hat{\imath}+3\hat{\jmath}| = \sqrt{18} = 3\sqrt{2}\).