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 :
- A \(49\)
- B \(42\)
- C \(50\)
- D \(45\)
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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