Let \(\vec{a}=2\hat{\imath}-3\hat{\jmath}+6\hat{k}\). Which of the following statements are correct?
(A) \(\vec{a}\cdot\hat{\imath} = 2\)
(B) The projection of \(\vec{a}\) on \(\hat{\jmath}\) is 3
(C) \(\vec{a}\times\hat{k} = -3\hat{\imath}-2\hat{\jmath}\)
(D) \(\vec{a}\times\hat{\imath} = -6\hat{\jmath}-3\hat{k}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
The scalar product with \(\hat{\imath}\), \(\hat{\jmath}\) or \(\hat{k}\) picks out a component:
\(\vec{a}\cdot\hat{\imath} = 2, \quad \vec{a}\cdot\hat{\jmath} = -3\)
For the vector products, use \(\hat{\imath}\times\hat{k} = -\hat{\jmath}\), \(\hat{\jmath}\times\hat{k} = \hat{\imath}\), \(\hat{\jmath}\times\hat{\imath} = -\hat{k}\), \(\hat{k}\times\hat{\imath} = \hat{\jmath}\):
\(\vec{a}\times\hat{k} = 2(\hat{\imath}\times\hat{k})-3(\hat{\jmath}\times\hat{k}) = -2\hat{\jmath}-3\hat{\imath}\)
\(\vec{a}\times\hat{\imath} = -3(\hat{\jmath}\times\hat{\imath})+6(\hat{k}\times\hat{\imath}) = 3\hat{k}+6\hat{\jmath}\)
So \(\vec{a}\cdot\hat{\imath} = 2\), the projection on \(\hat{\jmath}\) is \(-3\), \(\vec{a}\times\hat{k} = -3\hat{\imath}-2\hat{\jmath}\) and \(\vec{a}\times\hat{\imath} = 6\hat{\jmath}+3\hat{k}\).
Checking the statements
(A) \(\vec{a}\cdot\hat{\imath} = 2\) — Correct. It is the x-component of \(\vec{a}\).
(B) The projection on \(\hat{\jmath}\) is 3 — Incorrect. It is \(\vec{a}\cdot\hat{\jmath} = -3\).
(C) \(\vec{a}\times\hat{k} = -3\hat{\imath}-2\hat{\jmath}\) — Correct. \(-2\hat{\jmath}-3\hat{\imath}\) from the unit-vector products.
(D) \(\vec{a}\times\hat{\imath} = -6\hat{\jmath}-3\hat{k}\) — Incorrect. The signs are reversed: \(\vec{a}\times\hat{\imath} = 6\hat{\jmath}+3\hat{k}\).