Question

Determine the area of the parallelogram whose adjacent sides are formed by the vectors →A=ˆi−3ˆj+ˆk and →B=ˆi+ˆj+ˆk.

Answer 1

The area of the parallelogram formed by vectors
`\(\vec{A} = \hat{i} - 3\hat{j} + \hat{k}\)` and
`\(\vec{B} = \hat{i} + \hat{j} + \hat{k}\)` is
`\(√(20)\)` square units, calculated from the magnitude of their cross product.

The area A of a parallelogram formed by two vectors
`\(\vec{A}\)` and
`\(\vec{B}\)` is given by the magnitude of their cross product:

`\[ A = |\vec{A} * \vec{B}| \]`

The cross product
`\(\vec{A} * \vec{B}\)` is calculated as the determinant of the following matrix:

`\[ \vec{A} * \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \]`

Given vectors
`\(\vec{A} = \hat{i} - 3\hat{j} + \hat{k}\) and \(\vec{B} = \hat{i} + \hat{j} + \hat{k}\)`, we can find the cross product:

`\[ \vec{A} * \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]\[ \vec{A} * \vec{B} = (\hat{i})(-3 - 1) - (\hat{j})(1 - 1) + (\hat{k})(1 - 3) \]\[ \vec{A} * \vec{B} = -4\hat{i} - 2\hat{k} \]`

Now, find the magnitude of the cross product:

`\[ | \vec{A} * \vec{B} | = √((-4)^2 + 0^2 + (-2)^2) \]\[ | \vec{A} * \vec{B} | = √(16 + 4) \]\[ | \vec{A} * \vec{B} | = √(20) \]`

So, the area A of the parallelogram is
`\( √(20) \)` square units.