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Question: The two vectors 𝑖̂ + 𝑗̂ + 𝑘̂ and 3𝑖̂ − 𝑗̂ + 3𝑘̂ represent the two sides 𝑂𝐴 and 𝑂𝐵, respect...

The two vectors 𝑖̂ + 𝑗̂ + 𝑘̂ and 3𝑖̂ − 𝑗̂ + 3𝑘̂ represent the two sides 𝑂𝐴 and 𝑂𝐵, respectively of a ∆𝑂𝐴𝐵, where 𝑂 is the origin. The point 𝑃 lies on 𝐴𝐵 such that 𝑂𝑃 is a median. Find the area of the parallelogram formed by the two adjacent sides as 𝑂𝐴 and 𝑂𝑃

Answer

The area of the parallelogram formed by the two adjacent sides as OA and OP is 222\sqrt{2}.

Explanation

Solution

  1. Calculate the position vector of P (OP⃗\vec{OP}) as the midpoint of AB: OP⃗=OA⃗+OB⃗2\vec{OP} = \frac{\vec{OA} + \vec{OB}}{2}.
  2. Calculate the cross product OA⃗×OP⃗\vec{OA} \times \vec{OP}.
  3. The area of the parallelogram is the magnitude of the cross product: ∣OA⃗×OP⃗∣|\vec{OA} \times \vec{OP}|.

Given: OA⃗=i^+j^+k^\vec{OA} = \hat{i} + \hat{j} + \hat{k} OB⃗=3i^−j^+3k^\vec{OB} = 3\hat{i} - \hat{j} + 3\hat{k}

OP⃗=(i^+j^+k^)+(3i^−j^+3k^)2=4i^+4k^2=2i^+2k^\vec{OP} = \frac{(\hat{i} + \hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 3\hat{k})}{2} = \frac{4\hat{i} + 4\hat{k}}{2} = 2\hat{i} + 2\hat{k}

OA⃗×OP⃗=(i^+j^+k^)×(2i^+2k^)\vec{OA} \times \vec{OP} = (\hat{i} + \hat{j} + \hat{k}) \times (2\hat{i} + 2\hat{k}) =∣i^j^k^111202∣= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 0 & 2 \end{vmatrix} =i^(2−0)−j^(2−2)+k^(0−2)=2i^−2k^= \hat{i}(2-0) - \hat{j}(2-2) + \hat{k}(0-2) = 2\hat{i} - 2\hat{k}

Area = ∣2i^−2k^∣=22+02+(−2)2=4+4=8=22|2\hat{i} - 2\hat{k}| = \sqrt{2^2 + 0^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}.