JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 4)
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 :
$$\frac{7}{3}$$
3
$$\frac{10}{3}$$
$$\frac{8}{3}$$
Explanation
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} $
$$ \begin{aligned} \text { Now, } \overrightarrow{A B} \times \overrightarrow{A C} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{array}\right| \\\\ & =\hat{\mathbf{i}}(5)+\hat{\mathbf{j}}(-2-8)+\hat{\mathbf{k}}(4+6) \\\\ & =5 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \\\\ &\overrightarrow{O P} =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $$
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 $
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} $
$$ \begin{aligned} \text { Now, } \overrightarrow{A B} \times \overrightarrow{A C} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{array}\right| \\\\ & =\hat{\mathbf{i}}(5)+\hat{\mathbf{j}}(-2-8)+\hat{\mathbf{k}}(4+6) \\\\ & =5 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \\\\ &\overrightarrow{O P} =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $$
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 $
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