Question

The diagonals of a parallelogram are a =2i^3j^+k^ and b =2i^+4j^+k^ what is the area of the parallelogram?

Answer 1

Answer:

√5

Step-by-step explanation:

the diagonals of a parallelogram

Answer 2

Given:

The diagonals of a parallelogram are a =2i^3j^+k^ and b =2i^+4j^+k^.

To Find:

The area of the parallelogram.

Solution:

Let the parallelogram ABCD where AC and BD are diagonals of a parallelogram.

So,

AC=2i^3j^+k^ and BD = =2i^+4j^+k^

The magnitude of AC =

$\sqrt{ {2}^{2} + {3}^{2} + {1}^{2} } = \sqrt{14}$

The magnitude of BD =

$\sqrt{ {2}^{2} + {4}^{2} + {1}^{2} } = \sqrt{21}$

As we know,

Area of the parallelogram

$= \frac{1}{2} |magnitude \: of \: diagonal \: 1 \times magnitude \: of \: diagonal \: 2|$

$= \frac{1}{2} \sqrt{14} \times \sqrt{21}$

$= \frac{1}{2} | \: \sqrt{294} |$

$= \frac{7}{2} \sqrt{6}$

Henceforth, the area of a parallelogram is $\frac{7}{2} \sqrt{6}$