Area of a rectangle having vertices A- B- C and D with position vectors -12-4k- -12-4k- -12-4 k- and -12-4k- respectively is

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Vector Addition

Area of a rec...

Question

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

\frac{1}{2}

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Solution

The correct option is C $2$
The position vectors of vertices, $A, B, C$ , and $D$ of rectangle $A B C D$ are given as:

$\to O A = -^i + \frac{1}{2}^j + 4^k, \to O B =^i + \frac{1}{2}^j + 4^k, \to O C =^i - \frac{1}{2}^j + 4^k, \to O D = -^i - \frac{1}{2}^j + 4^k$

The adjacent sides $\to A B$ and $\to B C$ of the given rectangle are given as:

$\to A B = (1 + 1)^i + (\frac{1}{2} - \frac{1}{2})^j + (4 - 4)^k = 2^i$

$\to B C = (1 - 1)^i + (\frac{1}{2} - \frac{1}{2})^j + (4 - 4)^k = -^j$

$∴ \to A B \times \to B C = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & 0 & 0 0 & - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^k (- 2) = - 2^k$

$∣ ∣ \to A B \times \to A C ∣ ∣ = \sqrt{(- 2)^{2}} = 2$

Now, it is known that the area of a parallelogram whose adjacent sides are $\to a$ and $\to b$ is $∣ ∣ \to a \times \to b ∣ ∣$ .

Hence, the area of the given rectangle is $∣ ∣ \to A B \times \to B C ∣ ∣ = 2$ square units.
The correct answer is $C$ .

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Area of a rectangle having vertices A, B, C and D with position vectors −^i+12^j+4^k, ^i+12+4^k, ^i−12^j+4 ^k and −^i−12^j+4^k respectively is

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