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Distance Formula

If AB = 3, ...

Question

If $¯ ¯¯¯¯¯¯ ¯ A B = (3, - 2, 2), ¯ ¯¯¯¯¯¯ ¯ B C = (- 1, 0 - 2)$ are the adjacent sides of a parallelogram, then the obtuse angle between its diagonals is

A

\frac{π}{4}

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B

\frac{π}{2}

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C

\frac{π}{3}

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D

\frac{3 π}{4}

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Solution

The correct option is D $\frac{3 π}{4}$

$\Rightarrow$ $- - \to A B = (3, - 2, 2) = 3 ˆ i - 2 ˆ j + 2 ˆ k$
$\Rightarrow$ $- - \to B C = (- 1, 0, - 2) = - ˆ i - 2 ˆ k$
$\Rightarrow$ $- - \to A C = - - \to A B + - - \to B C = (3 ˆ i - 2 ˆ j + 2 ˆ k) + (- ˆ i - 2 ˆ k) = 2 ˆ i - 2 ˆ j$
$\Rightarrow$ $- - \to B D = - - \to B C + - - \to B A = - - \to B C + (- - - \to A B) = (- ˆ i - 2 ˆ k) + (- 3 ˆ i + 2 ˆ j - 2 ˆ k) = - 4 ˆ i + 2 ˆ j - 2 ˆ k$
$\Rightarrow$ $- - \to A C . - - \to B D = ∣ ∣ ∣ - - \to A C ∣ ∣ ∣ . ∣ ∣ ∣ - - \to B D ∣ ∣ ∣ . cos θ$
$\Rightarrow$ $(2) (- 4) + (- 2) (2) + (0) (- 4) = \sqrt{(2)^{2} + (- 2)^{2}} \times \sqrt{(- 4)^{2} + (2)^{2} + (- 4)^{2}} . cos θ$
$\Rightarrow$ $(- 8) + (- 4) + 0 = 2 \sqrt{2} \times 6 \times cos θ$
$\Rightarrow$ $- 12 = 2 \sqrt{2} \times 6 \times cos θ$
$\Rightarrow$ $cos θ = \frac{- 1}{\sqrt{2}} = cos (π - \frac{π}{4}) = cos (\frac{3 π}{4})$
$\Rightarrow$ $θ = \frac{3 π}{4}$

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