If O0,0,0, A2,3,−4, B3,4,−5 and cos∠OAB=p+2Q−3, then pQ+4=

A
$25$
B
49
C
30
D
9

Solution:

The given points are $O (0, 0, 0), A (2, 3, - 4), B (3, 4, - 5)$

Hence, the vectors are $\bar{A O} = - 2 i - 3 j + 4 k, \bar{A B} = i + j - k$

$\begin{matrix} \cos ∠ O A B = |\frac{- 2 - 3 - 4}{\sqrt{4 + 9 + 16} \sqrt{1 + 1 + 1}}| \\ = \frac{9}{\sqrt{29} \sqrt{3}} \\ = \sqrt{\frac{27}{29}} \end{matrix}$

Therefore, $p = 25, Q = 32$ and $\sqrt{25 (32 + 4)} = 5 \cdot 6 = 30$

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