If p→and q→are two diagonals of a quadrilateral such that p→-q→=p→.q→, p→=1,q→=2, then the area of quadrilateral is equal to :

|p→−q→|=p→⋅q→|p→|2+ |q→|2−2p→⋅q→=p→⋅q→ ⇒1+2=3p→⋅q→⇒p→⋅q→=1⇒p→q→cosθ=1∴ cosθ=12Area =12p→×q→=12 p→q→sinθ=12

If p→and q→are two diagonals of a quadrilateral such that p→-q→=p→.q→, p→=1,q→=2, then the area of quadrilateral is equal to :

|p→−q→|=p→⋅q→|p→|2+ |q→|2−2p→⋅q→=p→⋅q→ ⇒1+2=3p→⋅q→⇒p→⋅q→=1⇒p→q→cosθ=1∴ cosθ=12Area =12p→×q→=12 p→q→sinθ=12

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If $\vec{p}$ and $\vec{q}$ are two diagonals of a quadrilateral such that $|\vec{p} - \vec{q}| = \vec{p} . \vec{q}$ , $|\vec{p}| = 1, |\vec{q}| = \sqrt{2}$ , then the area of quadrilateral is equal to :

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Detailed Solution

$\begin{array}{l} | \vec{p} - \vec{q} | = \vec{p} \cdot \vec{q} \\ | \vec{p} |^{2} + | \vec{q} |^{2} - 2 \vec{p} \cdot \vec{q} = \vec{p} \cdot \vec{q} \Rightarrow 1 + 2 = 3 \vec{p} \cdot \vec{q} \\ \Rightarrow \vec{p} \cdot \vec{q} = 1 \Rightarrow |\vec{p}| |\vec{q}| cosθ = 1 \\ ∴ cosθ = \frac{1}{\sqrt{2}} \\ Area = \frac{1}{2} |\vec{p} \times \vec{q}| = \frac{1}{2} |\vec{p}| |\vec{q}| sinθ = \frac{1}{2} \end{array}$