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

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · 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}, 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

∠ BAC

where D is on the line segment BC , then

2 l^{2}

equals :

Answer: (D)

Let

A = 2 \hat{i} - 3 \hat{j} + 3 \hat{k}

B = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}

C = - \hat{i} + \hat{j} + 3 \hat{k}

A B = B - A = 5 \hat{j}

, so

A B = 5

A C = C - A = - 3 \hat{i} + 4 \hat{j}

, so

A C = 5

.

Since

A B = A C

, triangle

A B C

is isosceles with the two equal sides meeting at

A

, so the internal bisector of

∠ B A C

coincides with the median from

A

D

is the midpoint of

B C

D = \frac{B + C}{2} = \frac{( 2 - 1 ) i ^ + ( 2 + 1 ) j ^ + ( 3 + 3 ) k ^}{2} = \frac{1}{2} \hat{i} + \frac{3}{2} \hat{j} + 3 \hat{k} .

A D = D - A = (\frac{1}{2} - 2) \hat{i} + (\frac{3}{2} + 3) \hat{j} + (3 - 3) \hat{k} = - \frac{3}{2} \hat{i} + \frac{9}{2} \hat{j} .

l^{2} = ∣ A D ∣^{2} = \frac{9}{4} + \frac{81}{4} = \frac{90}{4} .

2 l^{2} = \frac{90}{2} = 45.

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option