Question

Find the area of a parallelogram whose diagonals 2□(→┬m-) □(→┬n ) and 4□(→┬m )- 5□(→┬n ) , where □(→┬m ) and □(→┬n ) are unit vectors forming an angle of 45°.

Answer 1

Answer on Question #46176 – Math – Vector Calculus

Problem.

Find the area of a parallelogram whose diagonals $2\square(\rightarrow_{\mathsf{T}}\mathsf{m}\cdot)\square(\rightarrow_{\mathsf{T}}\mathsf{n})$ and $4\square(\rightarrow_{\mathsf{T}}\mathsf{m})\cdot 5\square(\rightarrow_{\mathsf{T}}\mathsf{n})$, where $\square(\rightarrow_{\mathsf{T}}\mathsf{m})$ and $\square(\rightarrow_{\mathsf{T}}\mathsf{n})$ are unit vectors forming an angle of $45{}^{\circ}$.

Solution:

The magnitude of the cross product

$$(2 \vec {m} - \vec {n}) \times (4 \vec {m} - 5 \vec {n})$$

equal to the area of the parallelogram that this vectors span.

$$\begin{array}{l} (2 \vec {m} - \vec {n}) \times (4 \vec {m} - 5 \vec {n}) = 2 \vec {m} \times 4 \vec {m} + 2 \vec {m} \times (- 5 \vec {n}) + (- \vec {n}) \times 4 \vec {m} + (- \vec {n}) \times (- 5 \vec {n}) \\ = - 1 0 (\vec {m} \times \vec {n}) - 4 (\vec {n} \times \vec {m}) = - 1 0 (\vec {m} \times \vec {n}) + 4 (\vec {m} \times \vec {n}) = - 6 (\vec {m} \times \vec {n}), \\ \end{array}$$

as cross product of collinear vectors equals $\vec{0}$ and $\vec{m} \times \vec{n} = -\vec{n} \times \vec{m}$.

$$| - 6 (\vec {m} \times \vec {n}) | = 6 | \vec {m} \times \vec {n} | = 6 | \vec {m} | \cdot | \vec {n} | \cdot \sin (\vec {m}, \vec {n}) = 6 \cdot 1 \cdot 1 \cdot \sin 4 5 {}^ {\circ} = \frac {6}{\sqrt {2}} = 3 \sqrt {2}.$$

Answer: $3\sqrt{2}$

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