If $\vec a = 3\hat i-6\hat j +\hat k$ and $\vec b = 2\hat i-4\hat j+λ\hat k$ are such that $\vec a||\vec b$, then $3λ + 2 =$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
Given vectors:
\(\vec{a} = 3\hat{i} - 6\hat{j} + \hat{k}\)
\(\vec{b} = 2\hat{i} - 4\hat{j} + \lambda \hat{k}\)
Since \(\vec{a} \parallel \vec{b}\), there exists a scalar \(k\) such that:
\(\vec{b} = k \vec{a}\)
Equate components:
\[ \begin{cases} 2 = 3k \\ -4 = -6k \\ \lambda = k \end{cases} \]
From first equation:
\(k = \frac{2}{3}\)
From second equation:
\(-4 = -6k \Rightarrow k = \frac{4}{6} = \frac{2}{3}\)
Both equal \(k = \frac{2}{3}\), consistent.
From third equation:
\(\lambda = k = \frac{2}{3}\)
Calculate \(3\lambda + 2\):
\[ 3 \times \frac{2}{3} + 2 = 2 + 2 = 4 \]