Let \(\vec{a}=2\hat{\imath}+10\hat{\jmath}-11\hat{k}\).
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(|\vec{a}|\) | (I) | 30 |
| (B) | \(|-2\vec{a}|\) | (II) | 2 |
| (C) | Cosine of the angle that \(\vec{a}\) makes with the positive z-axis | (III) | 15 |
| (D) | The \(\hat{\jmath}\)-component of the vector of magnitude 3 in the direction of \(\vec{a}\) | (IV) | \(\displaystyle-\frac{11}{15}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A) \(|\vec{a}|\)
This matches (III) 15.
The magnitude is
\(\sqrt{4+100+121} = \sqrt{225} = 15\)
(B) \(|-2\vec{a}|\)
This matches (I) 30.
\(|-2\vec{a}| = |-2|\,|\vec{a}| = 2\times 15 = 30\)
(C) Cosine of the angle that \(\vec{a}\) makes with the positive z-axis
This matches (IV) \(-\frac{11}{15}\).
This is the third direction cosine, the z-component divided by the magnitude:
\( \displaystyle n = \frac{-11}{15} \)
(D) The \(\hat{\jmath}\)-component of the vector of magnitude 3 in the direction of \(\vec{a}\)
This matches (II) 2.
The required vector is 3 times the unit vector along \(\vec{a}\):
\( \displaystyle 3\times\frac{2\hat{\imath}+10\hat{\jmath}-11\hat{k}}{15} = \frac{2}{5}\hat{\imath}+2\hat{\jmath}-\frac{11}{5}\hat{k} \)
Its \(\hat{\jmath}\)-component is 2.