Q. Let P be a point in the plane of the vectors

→AB = 3î + ĵ − k̂ and →AC = î − ĵ + 3k̂

such that P is equidistant from the lines AB and AC. If

|→AP| = √5 / 2

then the area of the triangle ABP is :

Correct Answer: √30 / 4

Explanation

The vectors along the lines are

→AB = 3î + ĵ − k̂
→AC = î − ĵ + 3k̂

Since point P is equidistant from the lines AB and AC and lies in the plane of these vectors, the vector

→AP lies along the angle bisector of ∠BAC

The unit vectors along AB and AC are

|→AB| = √(3² + 1² + (−1)²) = √11
|→AC| = √(1² + (−1)² + 3²) = √11

So the unit vectors are

→u = (3î + ĵ − k̂)/√11
→v = (î − ĵ + 3k̂)/√11

The direction of the internal angle bisector is

→u + →v = (4î + 0ĵ + 2k̂)/√11

Hence

→AP = λ(4î + 2k̂)

Now,

|→AP| = λ√(4² + 2²) = λ√20

Given

λ√20 = √5 / 2

Solving,

λ = 1 / 4

Area of triangle ABP is

Area = 1/2 |→AB × →AP|

→AB × →AP =

| i  j  k |
| 3  1  −1 |
| 1  0  1/2 |

= (1/2)i − (5/2)j − i

|→AB × →AP| = √( (1/2)² + (5/2)² ) = √30 / 2

Therefore,

Area = 1/2 × √30 / 2 = √30 / 4

Hence, the required area is

√30 / 4

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