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JEE Mains · Maths · STD 12 - 10. vector algebra

The position vectors of the vertices \(A, B\) and \(C\) of a triangle are \(2 \hat{i}-3 \hat{j}+3 \hat{k}, \quad 2 \hat{i}+2 \hat{j}+3 \hat{k} \quad\) and \(-\hat{i}+\hat{j}+3 \hat{k}\) respectively. Let \(l\) denotes the length of the angle bisector \(\mathrm{AD}\) of \(\angle \mathrm{BAC}\) where \(\mathrm{D}\) is on the line segment \(\mathrm{BC}\), then \(2 l^2\) equals :

  1. A \(49\)
  2. B \(42\)
  3. C \(50\)
  4. D \(45\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(45\)

Step-by-step Solution

Detailed explanation

\(AB=5\) \(AC=5\) \(\therefore \mathrm{D}\) is midpoint of \(\mathrm{BC}\) \( \mathrm{D}\left(\frac{1}{2}, \frac{3}{2}, 3\right) \) \( \therefore l=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-3-\frac{3}{2}\right)^2+(3-3)^2} \) \( l=\sqrt{\frac{45}{2}} \) \( \therefore 2 l^2=45\)
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