A parallelogram is constructed on 5 a -2 b and a -3 b where - a -2 -2- b -3 and the angle between a and b is -4- If the magnitude of the longer diagonal is - where - Y is three digit number then - Y is

Position vector of diagonal =6 a b Length of diagonal =√(|6 a b|^2) =√(36|a|^2+|b|^2 12(a.b)) =√(36.8+9 12 × 2√(2)× 3 ×1√(2)) =√(225) P.V. of other diagona ...

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Standard XII

Mathematics

Applications of Dot Product

A parallelogr...

Question

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

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Solution

Position vector of diagonal $= 6 \to a - \to b$

Length of diagonal $= \sqrt{| 6 \to a - \to b |^{2}}$

$= \sqrt{36 | \to a |^{2} + | \to b |^{2} - 12 (\to a . \to b)}$

$= \sqrt{36.8 + 9 - 12 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}}$

$= \sqrt{225}$

P.V. of other diagonal $= 4 \to a + 5 \to b$

Length $= \sqrt{| 4 \to a + 5 \to b |^{2}}$

$= \sqrt{16 | \to a |^{2} + 25 | \to b |^{2} + 2.4 | \to a | .5 | \to b | . \frac{1}{\sqrt{2}}}$

$= \sqrt{593}$

So $α = 5, β = 9, γ = 3$

$β - α - γ = 1$

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A parallelogram is constructed on 5→a+2→b and →a−3→b where |→a|=2√2, |→b|=3 and the angle between →a and →b is π4. If the magnitude of the longer diagonal is √αβγ where αβγ is three digit number then (β−α−γ) is