Question

if the area of a parallelogram formed from the vectors A=î-2j+3k and B=3i-2j+k as adjacent sides

Answer 1

Answer:

The area of a parallelogram is 9.79 units.

Explanation:

Given that,

Vector $A=\hat{i}-2\hat{j}+3\hat{k}$

Vector $B =3\hat{i}-2\hat{j}+\hat{k}$

The area of a parallelogram is

$A = \vec{A}\times\vec{B}$

$A=\hat{i}-2\hat{j}+3\hat{k}\times3\hat{i}-2\hat{j}+\hat{k}$

$A = 4\hat{i}+8\hat{j}+4\hat{k}$

The magnitude is given by

$|A|=\sqrt{4^2+8^2+4^2}$

$|A|=9.79$

Hence, The area of a parallelogram is 9.79 units.

Answer 2

Explanation:

The area of a parallelogram is 9.79 units.

Given that,

Vector A=\hat{i}-2\hat{j}+3\hat{k}A=i^−2j^+3k^

Vector B =3\hat{i}-2\hat{j}+\hat{k}B=3i^−2j^+k^

The area of a parallelogram is

A = \vec{A}\times\vec{B}A=A×B

A=\hat{i}-2\hat{j}+3\hat{k}\times3\hat{i}-2\hat{j}+\hat{k}A=i^−2j^+3k^×3i^−2j^+k^

A = 4\hat{i}+8\hat{j}+4\hat{k}A=4i^+8j^+4k^

The magnitude is given by

|A|=\sqrt{4^2+8^2+4^2}∣A∣=42+82+42

|A|=9.79∣A∣=9.79

Hence, The area of a parallelogram is 9.79 units.