Question

find the area of the parallelogram whose adjacent sides are formed by the vector A = i` - 3j` +K` and B= i` + j` + k` ?​

Answer 1

Explanation:

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Answer 2

Given:

Adjacent sides of parallelogram is formed by vectors

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

$\vec{B} = \hat{i} + \hat{j} + \hat{k}$

To find:

Area of parallelogram.

Calculation:

The area of a parallelogram is given by the magnitude of vector product of the adjacent sides;

$\boxed{ \sf{area = | \vec{A} \times \vec{B}| }}$

$\sf{=>\vec{A}\times\vec{B} = \left[ \begin{array}{c c c} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{array} \right]}$

$\sf{ = > \vec{A} \times \vec{B} = \hat{i}( - 3 - 1) - \hat{j}(1 - 1) + \hat{k}(1 + 3)}$

$\sf{ = > \vec{A} \times \vec{B} = - 4\hat{i} - 0 \hat{j} + 4\hat{k}}$

$\sf{ = > \vec{A} \times \vec{B} = - 4\hat{i} + 4\hat{k}}$

$\sf{ = > | \vec{A} \times \vec{B} | = \sqrt{ {( - 4)}^{2} + {(4)}^{2} } }$

$\sf{ = > | \vec{A} \times \vec{B} | = \sqrt{ 16 + 16 } }$

$\sf{ = > | \vec{A} \times \vec{B} | = \sqrt{32 } }$

$\sf{ = > | \vec{A} \times \vec{B} | = 4\sqrt{2 } }$

So, final answer:

$\boxed{ \bf{ area = | \vec{A} \times \vec{B} | = 4\sqrt{2 } \: units}}$