Consider a ABC where A(1,3,2), B(-2,8,0) and C(3,6,7). If the angle bisector of BAC meets the line BC at D, then the length of the projection of the vector AD on the vector AC is :
- 372 38
- 19
- 392 38
- 382
Answer
(1) 372 38
Solution
AB= 9+25+4= 38, AC= 4+9+25= 38
the bisector divides BC in AB:AC=1:1 D = mid-point of BC = ( 12, 7, 72 )
[figure]
AD=- 12 i+4 j+ 32 k, AC=2 i+3 j+5 k
Projection = AD AC | AC |= -1+12+ 152 38
= 372 38
Answer: (1)