Let \(\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}\) and \(\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}\). If \(P\) is the point on the bisector of angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) such that \(|\overrightarrow{AP}|=\dfrac{\sqrt{5}}{2}\), then the area of \(\triangle APB\) is:

Question

The value of \( \hat{i} \cdot (\hat{k} \times \hat{j}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{j} \times \hat{i}) \) is _____

Answer 1

We are required to find the area of triangle APB, where point P lies on the angle bisector of vectors AB and AC, and |AP| = √5 / 2.

Step 1: Vectors AB and AC

Vector AB = (3, 1, −1)
Vector AC = (1, −1, 3)

Magnitude of AB:

|AB| = √(3² + 1² + (−1)²) = √11

Magnitude of AC:

|AC| = √(1² + (−1)² + 3²) = √11

Step 2: Unit vectors along AB and AC

Unit vector along AB = (3/√11, 1/√11, −1/√11)

Unit vector along AC = (1/√11, −1/√11, 3/√11)

Step 3: Direction of angle bisector

Angle bisector vector is the sum of the two unit vectors:

w = (4/√11, 0, 2/√11)

Step 4: Locate point P on the bisector

Let AP = t · w. Given |AP| = √5 / 2.

t · √[(4/√11)² + (2/√11)²] = √5 / 2

t · √(20/11) = √5 / 2

t = 1/2

Hence, vector AP is:

AP = (2/√11, 0, 1/√11)

Step 5: Area of triangle APB

Area of triangle APB is given by:

Area = 1/2 | AB × AP |

Compute the cross product:

AB × AP = (1/√11, 5/√11, −2/√11)

Magnitude of the cross product:

|AB × AP| = √[(1² + 25 + 4)/11] = √(30/11)

Therefore, area:

Area = 1/2 × √(30/11) = √30 / 4

Final Answer:

√30 / 4

Answer 2