Let P be a point in the plane of the vectors AB = 3 i + j - k and AC = i - j + 3 k such that P is equidistant from the lines AB and AC. If | AP| = 52, then the area of the triangle ABP is :
- 32
- 304
- 264
- 2
Answer
(2) 304
Solution
P is equidistant from lines AB and AC P lies on the bisector of BAC.
| AB| = | AC| = 11 bisector AB + AC = 4 i + 2 k 2 i + k
AP = 52 2 i + k 5 = i + 12 k
AB AP = vmatrix i & j & k \ 3 & 1 & -1 \ 1 & 0 & 12 vmatrix = 12 i - 52 j - k
| AB AP| = 14 + 254 + 1 = 302
area of ABP = 12 302 = 304
(P on the other bisector, AP AB - AC, gives area 54, which is not among the options.)
Answer: (2)