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Addition and Subtraction in Unit Vector Notation

If the positi...

Question

If the position vectors of the points $A, B, C, D$ are $(0, 2, 1), (3, 1, 1), (- 5, 3, 2), (2, 4, 1)$ respectively and if $- - \to P A + - - \to P B + - - \to P C + - - \to P D = \to 0,$ then the position vector of $P$ is

(0, \frac{5}{2}, \frac{5}{4})

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(\frac{5}{2}, \frac{5}{2}, \frac{5}{4})

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(\frac{5}{2}, 0, \frac{5}{4})

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(\frac{5}{2}, \frac{5}{4}, 0)

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Solution

The correct option is A $(0, \frac{5}{2}, \frac{5}{4})$
Given :
$- - \to P A + - - \to P B + - - \to P C + - - \to P D = \to 0 \Rightarrow (- - \to O A - - - \to O P) + (- - \to O B - - - \to O P) + (- - \to O C - - - \to O P) + (- - \to O D - - - \to O P) = \to 0$

$\Rightarrow 4 - - \to O P - (- - \to O A + - - \to O B + - - \to O C + - - \to O D) = \to 0 \Rightarrow - - \to O P = \frac{- - \to O A + - - \to O B + - - \to O C + - - \to O D}{4} \Rightarrow - - \to O P = (0, \frac{10}{4}, \frac{5}{4})$

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