Test Index

Vector Algebra Part 3

Section: Mathematics

Question : 1 of 72

Marks: +1, -0

\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}

\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}

|\overrightarrow{AP}|=\frac{\sqrt{5}}{2}

Solution:

Concept:

Point

P

is equidistant from lines

AB

and

AC

, so it lies on the angle bisector of the angle between

\overrightarrow{AB}

and

\overrightarrow{AC}

Explanation:

|\overrightarrow{AB}| = \sqrt{3^2+1^2+(-1)^2} = \sqrt{11}

and

|\overrightarrow{AC}| = \sqrt{1^2+(-1)^2+3^2} = \sqrt{11}

Since magnitudes are equal, the angle bisector direction is

\overrightarrow{AB}+\overrightarrow{AC} = (3+1)\hat{i}+(1-1)\hat{j}+(-1+3)\hat{k} = 4\hat{i}+2\hat{k}

Unit vector along bisector:

\hat{v} = \frac{4\hat{i}+2\hat{k}}{\sqrt{4^2+2^2}} = \frac{4\hat{i}+2\hat{k}}{2\sqrt{5}} = \frac{2\hat{i}+\hat{k}}{\sqrt{5}}

Given

|\overrightarrow{AP}| = \frac{\sqrt{5}}{2}

, so

\overrightarrow{AP} = |\overrightarrow{AP}|\hat{v} = \frac{\sqrt{5}}{2} \cdot \frac{2\hat{i}+\hat{k}}{\sqrt{5}} = \hat{i} + \frac{1}{2}\hat{k}

Area of

\triangle ABP = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AP}|

Compute cross product:

\overrightarrow{AB} \times \overrightarrow{AP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -1 \\ 1 & 0 & \frac{1}{2} \end{vmatrix} = \hat{i}\left(1\cdot\frac{1}{2} - (-1)\cdot0\right) - \hat{j}\left(3\cdot\frac{1}{2} - (-1)\cdot1\right) + \hat{k}\left(3\cdot0 - 1\cdot1\right) = \frac{1}{2}\hat{i} - \frac{5}{2}\hat{j} - \hat{k}

Magnitude:

\left|\frac{1}{2}\hat{i} - \frac{5}{2}\hat{j} - \hat{k}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{5}{2}\right)^2 + (-1)^2} = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{30}{4}} = \frac{\sqrt{30}}{2}

Area =

\frac{1}{2} \cdot \frac{\sqrt{30}}{2} = \frac{\sqrt{30}}{4}

Answer:

Option B:

\frac{\sqrt{30}}{4}

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