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

Given, the position vector of point P is : $ \overrightarrow{OP} = -\widehat{i} - 2\widehat{j} + 3\widehat{k} $ <br/><br/>Position vectors of points A, B, and C are : <br/><br/>$ \overrightarrow{OA} = -2\widehat{i} + \widehat{j} - 3\widehat{k} $ <br/><br/>$ \overrightarrow{OB} = 2\widehat{i} + 4\widehat{j} - 2\widehat{k} $ <br/><br/>$ \overrightarrow{OC} = -4\widehat{i} + 2\widehat{j} - \widehat{k} $ <br/><br/>Now, vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ can be calculated as : <br/><br/>$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} $ <br/><br/>$ = (2 + 2)\widehat{i} + (4 - 1)\widehat{j} - (-2 + 3)\widehat{k} $ <br/><br/>$ = 4\widehat{i} + 3\widehat{j} - \widehat{k} $ <br/><br/>$ \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} $ <br/><br/>$ = (-4 + 2)\widehat{i} + (2 - 1)\widehat{j} - (-1 + 3)\widehat{k} $ <br/><br/>$ = -2\widehat{i} + \widehat{j} - 2\widehat{k} $ <br/><br/>$$ \begin{aligned} \text { Now, } \overrightarrow{A B} \times \overrightarrow{A C} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{array}\right| \\\\ & =\hat{\mathbf{i}}(5)+\hat{\mathbf{j}}(-2-8)+\hat{\mathbf{k}}(4+6) \\\\ & =5 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \\\\ &\overrightarrow{O P} =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $$ <br/><br/>To find the projection of $ \overrightarrow{OP} $ onto $ \overrightarrow{AB} \times \overrightarrow{AC} $, we need to find the dot product between $ \overrightarrow{OP} $ and the normalized vector $ \overrightarrow{AB} \times \overrightarrow{AC} $. <br/><br/>First, find the magnitude of $ \overrightarrow{AB} \times \overrightarrow{AC} $ : <br/><br/>$ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{5^2 + (-10)^2 + 10^2} $ <br/><br/>$ = \sqrt{25 + 100 + 100} $ <br/><br/>$ = \sqrt{225} $ <br/><br/>$ = 15 $ <br/><br/>The projection of vector $ \overrightarrow{OP} $ onto a vector perpendicular to both $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ (which is $ \overrightarrow{AB} \times \overrightarrow{AC}$) is given by : <br/><br/>$ \frac{\overrightarrow{OP} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})}{| \overrightarrow{AB} \times \overrightarrow{AC} |} $ <br/><br/>= $ \frac{(-\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (5\hat{i} - 10\hat{j} + 10\hat{k})}{\sqrt{25 + 100 + 100}} $ <br/><br/>$ = \frac{-5 + 20 + 30}{15} $ <br/><br/>$ = \frac{45}{15} = 3 $