JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 8)
If the position vectors of the vertices A, B and C of a $$\Delta $$ ABC are respectively $$4\widehat i + 7\widehat j + 8\widehat k,$$ $$2\widehat i + 3\widehat j + 4\widehat k,$$ and $$2\widehat i + 5\widehat j + 7\widehat k,$$ then the position vectors of the point, where the bisector of $$\angle $$A meets BC is :
$${1 \over 2}\left( {4\widehat i + 8\widehat j + 11\widehat k} \right)$$
$${1 \over 3}\left( {6\widehat i + 11\widehat j + 15\widehat k} \right)$$
$${1 \over 3}\left( {6\widehat i + 13\widehat j + 18\widehat k} \right)$$
$${1 \over 4}\left( {8\widehat i + 14\widehat j + 19\widehat k} \right)$$
Explanation
Suppose angular bisector of A meets BC at D(x, , z)
Using angular bisector theorem,
$${{AB} \over {AC}}$$ = $${{BD} \over {DC}}$$
$${{BD} \over {DC}}$$ = $${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$$
= $${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$$ = $${6 \over 3}$$ = 2
_15th_April_Evening_Slot_en_8_1.png)
So, D(x, y, z) $$ \equiv $$ $$\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$$, $$\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$$
D(x, y, z) $$ \equiv $$ $$\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$$
Therefore, position vector of point p = $${1 \over 3}$$ (6i + 13j + 18k)
Using angular bisector theorem,
$${{AB} \over {AC}}$$ = $${{BD} \over {DC}}$$
$${{BD} \over {DC}}$$ = $${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$$
= $${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$$ = $${6 \over 3}$$ = 2
So, D(x, y, z) $$ \equiv $$ $$\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$$, $$\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$$
D(x, y, z) $$ \equiv $$ $$\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$$
Therefore, position vector of point p = $${1 \over 3}$$ (6i + 13j + 18k)
Comments (0)
