If the vectors \(\rm \vec{a}, \vec{b}\) are collinear and \(\rm \vec{a} = 2\hat{i} +6\hat{j} - 3\hat{k}\) and |b| = 14, then \(\rm \vec{b}\) is equal to?

​Concept:

Let \({\rm{\vec a}} = {\rm{x\;\vec i}} + {\rm{y\;\vec j}} + {\rm{z\;\vec k}}\)

Magnitude of the vector of a = \(\left| {{\rm{\vec a}}} \right| = {\rm{\;}}\sqrt {{{\rm{x}}^2} + {\rm{\;}}{{\rm{y}}^2} + {{\rm{z}}^2}} \)

Collinear vectors: Two vectors are collinear if they lie on the same line or parallel lines.

If \(\rm \vec{a}\) and \(\rm \vec{b}\) are collinear vectors then \(\rm \vec{b} = λ \vec{a}\)​

Calculation:

Given:

vectors \(\rm \vec{a}, \vec{b}\) are collinear,

Therefore, \(\rm \vec{b} = λ \vec{a}\)

⇒ \(\rm \vec{b} = λ (2\hat{i} +6\hat{j} - 3\hat{k})\)

Given: magnitude of b = |b| = 14

⇒ \(|λ (2\hat{i} +6\hat{j} - 3\hat{k})| = 14\)

⇒ \(λ| (2\hat{i} +6\hat{j} - 3\hat{k})| = 14\)

⇒ \(λ \times \sqrt {2^2+6^2+(-3)^2} = 14\)

⇒ 7λ = 14

∴ λ = 2

Now, \(\rm \vec{b} = λ \vec{a} = λ (2\hat{i} +6\hat{j} - 3\hat{k}) = 2 (2\hat{i} +6\hat{j} - 3\hat{k})\)

Hence \(\rm \vec{b}= (4\hat{i} +12\hat{j} - 6\hat{k})\)