Let O be the origin and the position vector of the point P be -i-2j+3k.

Question

Let O be the origin and the position vector of the point P be $-\hat{i}-2\hat{j}+3\hat{k}$. If the position vectors of the points A, B and C are $-2\hat{i}+\hat{j}-3\hat{k},2\hat{i}+4\hat{j}-2\hat{k}$ and $-4\hat{i}+2\hat{j}-\hat{k}$ respectively, then the projection of the vector $\overset{	o \leftarrow }{OP}$ on a vector perpendicular to the vectors $\overset{	o \leftarrow }{AB}$ and $\overset{	o \leftarrow }{AC}$ is :

Accepted Answer

Given, the position vector of point P is :
$\overset{	o \leftarrow }{OP}=-\hat{i}-2\hat{j}+3\hat{k}$

Position vectors of points A, B, and C are :

$\overset{	o \leftarrow }{OA}=-2\hat{i}+\hat{j}-3\hat{k}$

$\overset{	o \leftarrow }{OB}=2\hat{i}+4\hat{j}-2\hat{k}$

$\overset{	o \leftarrow }{OC}=-4\hat{i}+2\hat{j}-\hat{k}$

Now, vectors $\overset{	o \leftarrow }{AB}$ and $\overset{	o \leftarrow }{AC}$ can be calculated as :

$\overset{	o \leftarrow }{AB}=\overset{	o \leftarrow }{OB}-\overset{	o \leftarrow }{OA}$

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

$=4\hat{i}+3\hat{j}-\hat{k}$

$\overset{	o \leftarrow }{AC}=\overset{	o \leftarrow }{OC}-\overset{	o \leftarrow }{OA}$

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

$=-2\hat{i}+\hat{j}-2\hat{k}$

$\begin{gathered} \overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC} \quad  =|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 4 \quad  3 \quad  1 \ -2 \quad  1 \quad  2\end{array}| \ =\hat{i}(5)+\hat{j}(-2-8)+\hat{k}(4+6) \ =5\hat{i}-10\hat{j}+10\hat{k} \ \overset{	o \leftarrow }{OP}=-\hat{i}-2\hat{j}+3\hat{k} \end{gathered}$

To find the projection of $\overset{	o \leftarrow }{OP}$ onto $\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC}$, we need to find the dot product between $\overset{	o \leftarrow }{OP}$ and the normalized vector $\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC}$.

First, find the magnitude of $\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC}$ :

$|\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC}|=\sqrt{{5}^{2}+(-10{)}^{2}+{10}^{2}}$

$=\sqrt{25+100+100}$

$=\sqrt{225}$

$=15$

The projection of vector $\overset{	o \leftarrow }{OP}$ onto a vector perpendicular to both $\overset{	o \leftarrow }{AB}$ and $\overset{	o \leftarrow }{AC}$ (which is $\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC}$) is given by :

$\frac{\overset{	o \leftarrow }{OP}\cdot (\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC})}{|\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC}|}$

= $\frac{(-\hat{i}-2\hat{j}+3\hat{k})\cdot (5\hat{i}-10\hat{j}+10\hat{k})}{\sqrt{25+100+100}}$

$=\frac{-5+20+30}{15}$

$=\frac{45}{15}=3$

Solutions · 1