Let P be a point in the plane of the vectors AB=3i+j-k and AC=i-j+3k such that P is equidi…

Question

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

Accepted Answer

Point $P$ is equidistant from the lines $AB$ and $AC$. Geometrically, this means point $P$ must lie on the angle bisector of the angle formed by vectors $AB$ and $AC$.

First, we find the unit vectors along $AB$ and $AC$ :

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

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

Since the magnitudes are equal, the angle bisector $v$ is simply the sum of the two vectors:

$\begin{gathered} v=AB+AC=(3+1)\hat{i}+(1-1)\hat{j}+(-1+3)\hat{k}=4\hat{i}+2\hat{k} \end{gathered}$

The vector $AP$ will be in the direction of the unit vector $\hat{v}$ :

$\hat{v}=\frac{4\hat{i}+2\hat{k}}{\sqrt{{4}^{2}+{2}^{2}}}=\frac{4\hat{i}+2\hat{k}}{\sqrt{20}}=\frac{4\hat{i}+2\hat{k}}{2\sqrt{5}}=\frac{2\hat{i}+\hat{k}}{\sqrt{5}}$

Determine Vector $AP$

Given that $|AP|=\frac{\sqrt{5}}{2}$, we can write:

$AP=|AP|\hat{v}=\frac{\sqrt{5}}{2}(\frac{2\hat{i}+\hat{k}}{\sqrt{5}})=\frac{1}{2}(2\hat{i}+\hat{k})=\hat{i}+\frac{1}{2}\hat{k}$

Calculate the Area of Triangle $ABP$

The area of a triangle formed by two vectors $AB$ and $AP$ is given by the formula:

$=\frac{1}{2}|\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AP}|$

First, compute the cross product $AB	imes AP$ :

$\begin{gathered} \overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AP}=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 3 \quad  1 \quad  -1 \ 1 \quad  0 \quad  \frac{1}{2}\end{array}| \ =\hat{i}(\frac{1}{2}-0)-\hat{j}(\frac{3}{2}-(-1))+\hat{k}(0-1) \ =\frac{1}{2}\hat{i}-\frac{5}{2}\hat{j}-\hat{k} \ |\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AP}|=\sqrt{{(\frac{1}{2})}^{2}+{(-\frac{5}{2})}^{2}+(-1{)}^{2}}=\sqrt{\frac{1}{4}+\frac{25}{4}+1}=\sqrt{\frac{26}{4}+\frac{4}{4}}=\sqrt{\frac{30}{4}}=\frac{\sqrt{30}}{2} \end{gathered}$

Finally, the area of the triangle is:

Area $=\frac{1}{2}(\frac{\sqrt{30}}{2})=\frac{\sqrt{30}}{4}$

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