If the sides AB and AC of \(\triangle ABC\) are represented by vectors \(\hat{j}+\hat{k}\) and \(3\hat{i}-\hat{j}+4\hat{k}\) respectively, then the length of the median through A on BC is:
A. \(2\sqrt{2}\) units
B. \(\sqrt{18}\) units
C. \(\frac{\sqrt{34}}{2}\) units
D. \(\frac{\sqrt{48}}{2}\) units
Official Solution
Correct Answer: \(\frac{\sqrt{34}}{2}\) units
Explanation:
The position vector of the midpoint \(D\) is the average of the position vectors of \(B\) and \(C\) relative to \(A\):\[\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}\]
\[\vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}\]
The length of the median is the magnitude of the vector \(\vec{AD}\), denoted as \(|\vec{AD}|\):\[|\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(0\right)^2 + \left(\frac{5}{2}\right)^2}\]
\[|\vec{AD}| = \frac{\sqrt{34}}{2}\]
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