The vertices of a triangle ABC are $A (- 1, 2, - 3), B (5, 0, - 6), C (0, 4, - 1) .$ Then the direction ratios of the external bisector of $∠ B A C$ are :

11, 20, 21

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- 11, 20, 20

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- 11, 20, 23

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Solution

The correct option is B $- 11, 20, 23$
Given A(−1,2,−3),B(5,0,−6),C(0,4,−1)
$\to A B = \to a = \to B - \to A$
$\to A B = \to a = 5^i - 6^k +^i - 2^j + 3^k$
$\to A B = \to a = 6^i - 2^j - 3^k$
$\to A C = \to c = \to C - \to A$
$\to A C = \to c = 4^j -^k +^i - 2^j + 3^k$
$\to A C = \to c =^i + 2^j + 2^k$
external angle bisector $=^c -^a$
external angle bisector $= \frac{^i + 2^j + 2^k}{\sqrt{1^{2} + 2^{2} + 2^{2}}} - \frac{6^i - 2^j - 3^k}{\sqrt{6^{2} + 2^{2} + 3^{2}}}$
external angle bisector $= \frac{^i + 2^j + 2^k}{\sqrt{9}} - \frac{6^i - 2^j - 3^k}{\sqrt{49}}$
external angle bisector $= \frac{^i + 2^j + 2^k}{3} - \frac{6^i - 2^j - 3^k}{7}$
external angle bisector $= \frac{^i}{3} + \frac{2^j}{3} + \frac{2^k}{3} - \frac{6^i}{7} - \frac{^j}{7} - \frac{3^k}{7}$
external angle bisector $= \frac{7^i - 18^i}{21} + \frac{14^j + 6^j}{21} + \frac{14^k + 9^k}{21}$
external angle bisector $= \frac{- 11^i}{21} + \frac{20^j}{21} + \frac{23^k}{21}$
direction ratio will be $- 11, 20, 23$

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Similar questions

A (- 1, 2, - 3), B (5, 0, - 6)

C (0, 4, - 1)

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A (- 1, 2, 3), B (5, 0, - 6), C (0, 4, - 1)

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