The position vectors of the vertices A, B and C of a triangle are , and respectively. Let denotes the length of the angle bisector AD of where D is on the line segment BC, then equals : [2024]

Question

The position vectors of the vertices A, B and C of a triangle are $2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ , $2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $- \hat{i} + \hat{j} + 3 \hat{k}$ respectively. Let $l$ denotes the length of the angle bisector AD of $∠ B A C$ where D is on the line segment BC, then $2 l^{2}$ equals : [2024]

Smart Achievers · Accepted Answer

(3)

In , $△ A B C$

Position vector of $A = 2 \hat{i} - 3 \hat{j} + 3 \hat{k}$

Position vector of $B = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$

Position vector of $C = - \hat{i} + \hat{j} + 3 \hat{k}$

Coordinates of the triangle ABC are A(2, –3, 3), B(2, 2, 3) and C(–1, 1, 3).

AD is the angle bisector of $∠ B A C$ .

$\Rightarrow B D : D C = A B : A C \Rightarrow D$ is the mid-point of the side BC.

Let the coordinates of D be (x, y, z).

By mid-point formula, $x = \frac{2 - 1}{2}, y = \frac{2 + 1}{2}, z = \frac{3 + 3}{2}$

$\Rightarrow x = \frac{1}{2}, y = \frac{3}{2} and z = 3$

So, $A D^{2} = {(2 - \frac{1}{2})}^{2} + {(- 3 - \frac{3}{2})}^{2} + {(3 - 3)}^{2} = \frac{90}{4}$

$\Rightarrow l^{2} = A D^{2} = \frac{90}{4}$ $∴ 2 l^{2} = 2 \times \frac{90}{4} = 45$

Solution

1 50

2 49

3 45

4 42

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