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Scalar Triple Product

For non-zero ...

Question

For non-zero vectors $\to p, \to q$ and $\to r,$ let $(\to p \times \to q) \times \to r + (\to q \cdot \to r) \to q = (x^{2} + y^{2}) \to q + (14 - 4 x - 6 y) \to p$ and $(\to r \cdot \to r) \to p = \to r$ where $\to p$ and $\to q$ are non-collinear, and $x$ and $y$ are scalars. Then the value of $(x + y)$ is

A

0

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B

5

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C

10

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D

1

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Solution

The correct option is B $5$
$(\to p \cdot \to r) \to q - (\to q \cdot \to r) \to p + (\to q \cdot \to r) \to q = (x^{2} + y^{2}) \to q + (14 - 4 x - 6 y) \to p$
$∴ \to p \cdot \to r + \to q \cdot \to r = x^{2} + y^{2} \dots (1)$
and $- \to q \cdot \to r = 14 - 4 x - 6 y \dots (2)$
From $(1) + (2)$
$\to p \cdot \to r = x^{2} + y^{2} - 4 x - 6 y + 14 \dots (3)$

Since $(\to r \cdot \to r) \to p = \to r,$
taking dot product with $\to r,$ we get
$(\to r \cdot \to r) (\to p \cdot \to r) = (\to r \cdot \to r)$
$\Rightarrow \to p \cdot \to r = 1$

$∴$ From $(3),$
$x^{2} + y^{2} - 4 x - 6 y + 14 = 1 \Rightarrow {(x - 2)}^{2} + {(y - 3)}^{2} = 0 \Rightarrow x = 2, y = 3$
Hence, $x + y = 5$

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