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Vector Algebra Part 2

Section: Mathematics

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Question : 19 of 100

Marks: +1, -0

Let

O

be the origin and the position vector of the point

P

-\hat{i}-2\hat{j}+3\hat{k}

. If the position vectors of the

A, B

C

-2\hat{i}+\hat{j}-3\hat{k}, 2\hat{i}+4\hat{j}-2\hat{k}

-4\hat{i}+2\hat{j}-\hat{k}

respectively, then the projection of the vector

\overrightarrow{OP}

on a vector perpendicular to the vectors

\overrightarrow{AB}

\overrightarrow{AC}

[10-Apr-2023 shift 1]

$\;\frac{10}{3}$
$\;\frac{8}{3}$
$\;\frac{7}{3}$
3

Solution:

Position vector of the point

P(-1,-2,3), A(-2,1,-3) B(2,4,-2)

, and

C(-4,2,-1)

Then

\overrightarrow{OP} \cdot \; \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{\left|\overrightarrow{AB} \times \overrightarrow{AC})\right|}

\;\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{vmatrix}

\; = \hat{i}(5)-\hat{j}(8+2)+\hat{k}(4+6)

\; = 5\hat{i}-10\hat{j}+10\hat{k}

Now

\;\overrightarrow{OP} \cdot \; \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{\left|\overrightarrow{AB} \times \overrightarrow{AC})\right|} = (-\hat{i} - 2\hat{j} + 3\hat{k}) \cdot \frac{\hat{(\hat{i} - 10\hat{j} + 10\hat{k})}}{\sqrt{(5)^2+(-10)^2+(10)^2}}

\; = \; \frac{-5+20+30}{\sqrt{25+100+100}}

\; = \; \frac{45}{\sqrt{225}} = \; \frac{45}{15} = 3

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