AP EAMCET · Maths · Three Dimensional Geometry

$A (- 1, 2, - 3), B (5, 0, - 6), C (0, 4, - 1)$ are the vertices of a triangle $A B C$ . The direction cosines of internal bisector of $∠ B A C$ are

A $\frac{25}{\sqrt{714}}, \frac{8}{\sqrt{714}}, \frac{- 5}{\sqrt{714}}$
B $\frac{25}{\sqrt{714}}, \frac{8}{\sqrt{714}}, \frac{5}{\sqrt{714}}$
C $\frac{5}{\sqrt{74}}, \frac{6}{\sqrt{74}}, \frac{8}{\sqrt{74}}$
D $\frac{- 5}{\sqrt{74}}, \frac{6}{\sqrt{74}}, \frac{- 8}{\sqrt{74}}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{25}{\sqrt{714}}, \frac{8}{\sqrt{714}}, \frac{5}{\sqrt{714}}$

Step-by-step Solution

Detailed explanation

Given pointsA(-1,2,-3), B(5,0,-6), C(0,4,-1) Let, the vector AD is the internal bisector of ∠BAC, point D lie on the side BC. Now, AB=-1-52+22+-3+62=36+4+9=7 and, AC=-1-02+2-42+-3+12=1+4+4=3 We know that, the bisector of angle ∠BAC, divides the side BC…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A(-1,2,-3), B(5,0,-6), C(0,4,-1) are the vertices of a triangle ABC. The direction cosines of internal bisector of ∠BAC are

Step-by-step Solution

$A (- 1, 2, - 3), B (5, 0, - 6), C (0, 4, - 1)$ are the vertices of a triangle $A B C$ . The direction cosines of internal bisector of $∠ B A C$ are