First slide

Dot product or scalar product of vectors

Question

$A parallelogram is constructed on 5 \vec{a} + 2 \vec{b} and \vec{a} - 3 \vec{b} as adjacent sides, where | \vec{a} | = 2 \sqrt{2}$
$and | \vec{b} | = 3 & angle between \vec{a} & \vec{b} is \frac{π}{4} . If the magnitude of longer diagonal is \sqrt{α β γ}$
$where α β γ is three digit number then α + β - γ is equal to$

Moderate

A

-1
B

1
C

11
D

7

Solution

$Given that | \vec{a} | = 2 \sqrt{2}, | \vec{b} | = 3 and angle between \vec{a} and \vec{b} is \frac{π}{4}$
$\begin{array}{l} The diagonals of the parallelogram are \\ \vec{p} = 5 \vec{a} + 2 \vec{b} + \vec{a} - 3 \vec{b} = 6 \vec{a} - \vec{b} and \vec{q} = 4 \vec{a} + 5 \vec{b} \\ | \vec{p} | = | \vec{6} \vec{a} - \vec{b} | = \sqrt{36 | \vec{a} |^{2} + | \vec{b} |^{2} - 12 (\vec{a} \cdot \vec{b})} \\ = \sqrt{36 \times 8 + 9 - 12 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}} = 15 \end{array}$
$\begin{array}{l} And | \vec{q} | = | 4 \vec{a} + 5 \vec{b} | = \sqrt{16 | \vec{a} |^{2} + 25 | \vec{b} | + 40 (\vec{a \cdot} \vec{b})} \\ = \sqrt{16 \times 8 + 25 \times 9 + 40 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}} = \sqrt{593} \\ ∴ Magnitude of longer diagonal = \sqrt{593} = \sqrt{α β γ} (given) \\ ∴ α + β - γ = 5 + 9 - 3 = 11 \end{array}$

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