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Applications of Cross Product

Area of a rec...

Question

Area of a rectangle having vertices A, B, C and D with position vectors $- ˆ i + \frac{1}{2} ˆ j + 4 ˆ k, ˆ i + \frac{1}{2} ˆ j + 4 ˆ k, ˆ i - \frac{1}{2} ˆ j + 4 ˆ k$ and $- ˆ i - \frac{1}{2} ˆ j + 4 ˆ k$ , respectively is

A

\frac{1}{2}

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B

1

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C

2

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D

4

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Solution

The correct option is B $2$
$\begin{matrix} - - \to O A = - ˆ i + \frac{1}{2} ˆ j + 4 ˆ k = - 1 ˆ i + \frac{1}{2} ˆ j + 4 ˆ k - - \to O B = ˆ i + \frac{1}{2} ˆ j + 4 ˆ k = 1 ˆ i + \frac{1}{2} ˆ j + 4 ˆ k - - \to O C = ˆ i - \frac{1}{2} ˆ j + 4 ˆ k = 1 ˆ i - \frac{1}{2} ˆ j + 4 ˆ k - - \to O D = - ˆ i - \frac{1}{2} ˆ j + 4 ˆ k = - 1 ˆ i - \frac{1}{2} ˆ j + 4 ˆ k \end{matrix}$
Since rectangle is also a parallelogram
Area of rectangle ABCD $= ∣ ∣ ∣ - - \to A B \times - - \to B C ∣ ∣ ∣$
$\begin{matrix} - - \to A B = - - \to O B - - - \to O A = (1 ˆ i + \frac{1}{2} ˆ j + 4 ˆ k) - (- 1 ˆ i + \frac{1}{2} ˆ j + 4 ˆ k) = (1 - (- 1) ˆ i + (\frac{1}{2} - \frac{1}{2}) ˆ j + (4 - 4)) ˆ k = 2 ˆ i + 0 ˆ j + 0 ˆ k - - \to B C = - - \to O C - - - \to O B = (1 ˆ i - \frac{1}{2} ˆ j + 4 ˆ k) - (1 ˆ i + \frac{1}{2} ˆ j + 4 ˆ k) = ((1 - 1) ˆ i + (- \frac{1}{2} - \frac{1}{2}) ˆ j + (4 - 4)) ˆ k = 0 ˆ i - 1 ˆ j + 0 ˆ k ∣ ∣ ∣ - - \to A B \times - - \to B C ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i ˆ j ˆ k 200 0 - 10 \end{matrix} ∣ ∣ ∣ ∣ ∣ = ˆ i (0 \times 0 - (- 1) \times 0) - ˆ j (2 \times 0 - 0 \times 0) + ˆ k (2 \times - 1 - 0 \times 0) = ˆ i (0 - 0) - ˆ j (0 - 0) + ˆ k (- 2 - 0) = 0 ˆ i - 0 ˆ j - 2 ˆ k \end{matrix}$
Now,
Magnitude of
$\begin{matrix} - - \to A B \times - - \to B C = \sqrt{0^{2} + 0^{2} + {(- 2)}^{2}} ∣ ∣ ∣ - - \to A B \times - - \to B C ∣ ∣ ∣ = \sqrt{4} = 2 \end{matrix}$
Therefore,
area of rectangle ABCD= $∣ ∣ ∣ - - \to A B \times - - \to B C ∣ ∣ ∣ = 2$

Hence, Option $C$ is the correct answer.

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