Question

If and $2 \mathbf { i } + 5 \mathbf { j } + 7 \mathbf { k }$ are the position vectors of the vertices A, B and C respectively of triangle ABC. The position vector of the point where the bisector of angle A meets BC is

• A. $\frac { 1 } { 3 } ( 6 \mathbf { i } + 13 \mathbf { j } + 18 \mathbf { k } )$
• B.
• C. $\frac { 1 } { 3 } ( - 6 \mathbf { i } - 8 \mathbf { j } - 9 \mathbf { k } )$
• D. $\frac { 2 } { 3 } ( - 6 \mathbf { i } - 12 \mathbf { j } + 8 \mathbf { k } )$

Answer 1

Answer: (A)

$\frac { 1 } { 3 } ( 6 \mathbf { i } + 13 \mathbf { j } + 18 \mathbf { k } )$

Answer 2

Let the bisector of angle A meets BC at D, then AD divides BC in the ratio AB: AC

∴ Position vectors of D

=$\frac { | \overrightarrow { A B } | ( 2 \mathbf { i } + 5 \mathbf { j } + 7 \mathbf { k } ) + | \overrightarrow { A C } | ( 2 \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k } ) } { | \overrightarrow { A B } | + | \overrightarrow { A C } | }$

Here, and $| \overrightarrow { A C } | = - 2 \mathbf { i } - 2 \mathbf { j } - \mathbf { k } \mid = 3$

∴ Position vector of $D = \frac { 6 ( 2 \mathbf { i } + 5 \mathbf { j } + 7 \mathbf { k } ) + 3 ( 2 \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k } ) } { 6 + 3 }$

$= \frac { 18 \mathbf { i } + 39 \mathbf { j } + 54 \mathbf { k } ) } { 9 }$ $= \frac { 1 } { 3 } ( 6 \mathbf { i } + 13 \mathbf { j } + 18 \mathbf { k } )$.