QJEEM23Ma1e79: JEE Main 2023 Physics - Class 11 JEEM23Ma1e79 Solution: Let O be the origin and the position vector of the point P be − …

Given, the position vector of point P is : $\overrightarrow{OP} = -\widehat{i} - 2\widehat{j} + 3\widehat{k}$

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

$\overrightarrow{OA} = -2\widehat{i} + \widehat{j} - 3\widehat{k}$

$\overrightarrow{OB} = 2\widehat{i} + 4\widehat{j} - 2\widehat{k}$

$\overrightarrow{OC} = -4\widehat{i} + 2\widehat{j} - \widehat{k}$

Now, vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ can be calculated as :

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$

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

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

$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}$

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

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

$$\begin{array}{rl}\text{ Now, }\overrightarrow{AB}\times \overrightarrow{AC}& =\left|\begin{array}{ccc}\hat{\mathbf{i}}& \hat{\mathbf{j}}& \hat{\mathbf{k}}\\ 4& 3& 1\\ -2& 1& 2\end{array}\right|\\ \\ & =\hat{\mathbf{i}}(5)+\hat{\mathbf{j}}(-2-8)+\hat{\mathbf{k}}(4+6)\\ \\ & =5\hat{\mathbf{i}}-10\hat{\mathbf{j}}+10\hat{\mathbf{k}}\\ \\ & \overrightarrow{OP}=-\hat{\mathbf{i}}-2\hat{\mathbf{j}}+3\hat{\mathbf{k}}\end{array}$$

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 $\overrightarrow{AB} \times \overrightarrow{AC}$ .

First, find the magnitude of $\overrightarrow{AB} \times \overrightarrow{AC}$ :

$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{5^2 + (-10)^2 + 10^2}$

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

$= \sqrt{225}$

$= 15$

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 :

$\frac{\overrightarrow{OP} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})}{| \overrightarrow{AB} \times \overrightarrow{AC} |}$

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

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

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