Let P be a point in the plane of the vectors AB=3i+j-k and AC=i-j+3k such that P is equidistant from the lines AB and AC . If | AP |= 52, then the area of the triangle ABP is:

Question

Accepted Answer

Correct answer: B. (b) Since P is equidistant from lines AB and AC through point A, it lies along the angle bisector direction. The unit vectors are u_ AB= 3i+j-k 11 and u_ AC= i-j+3k 11. The bisector direction is proportional to 2i+k, so AP= (2i+k ). From | AP |= 52 : | | 5= 52, giving = 12. Thus AP=i+ 12k. Area of triangle ABP= 12 | AB AP |. Computing: AB AP= 12i- 52j-k. | AB AP |= 14+ 254+1= 304= 302. Area = 304.

$Let P be a point in the plane of the vectors AB = 3 i + j - k and AC = i - j + 3 k such that P is equidistant from the lines AB and AC . If | AP | = \frac{5}{2}, then the area of the triangle ABP is:$

Solution

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