Discussion
The position vectors of the points A and B are \(\vec{a} = 2\hat{i} - \lambda\hat{j} + 5\hat{k}\) and \(\vec{b} = \mu\hat{i} + 7\hat{j} + 3\hat{k}\) respectively. If the position vector of the mid-point of the line segment AB is \(\vec{c} = 3\hat{i} + 2\hat{j} + 4\hat{k}\), then the value of \(\lambda + \mu\) is equal to
Answer: B
\(\lambda + \mu = 7\).
The mid-point has position vector \(\frac{\vec{a} + \vec{b}}{2}\). Compare components with \(3\hat{i} + 2\hat{j} + 4\hat{k}\):
\[ \begin{aligned} \tfrac{2 + \mu}{2} &= 3 \Rightarrow \mu = 4 \\ \tfrac{-\lambda + 7}{2} &= 2 \Rightarrow \lambda = 3 \\ \tfrac{5 + 3}{2} &= 4 \text{ (consistent)} \end{aligned} \]
So \(\lambda + \mu = 3 + 4 = 7\). Taking the j-component of \(\vec{a}\) as \(+\lambda\) instead of \(-\lambda\) gives \(\lambda = -3\) and a sum of 1, which is not offered.
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