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Chapter 5: Problem 25

Find the area of the parallelogram that has the vectors as adjacent sides. $$\mathbf{u}=\mathbf{j}, \quad \mathbf{v}=\mathbf{j}+\mathbf{k}$$

Short Answer

Expert verified
The area of the parallelogram that has the vectors \(\mathbf{u}=\mathbf{j}\) and \(\mathbf{v}=\mathbf{j} + \mathbf{k}\) as adjacent sides is 1.

Step by step solution

01

Interpret the Given Vectors

The vectors given in the problem \(\mathbf{u} = \mathbf{j}\) and \(\mathbf{v} = \mathbf{j} + \mathbf{k}\) can be rewritten in component form for easier computation. Thus, \(\mathbf{u} = <0, 1, 0>\) and \(\mathbf{v} = <0, 1, 1>\). Since these vectors are in 3D, cross product can be calculated.
02

Compute the Cross Product

The cross product of two vectors can be computed using the determinant of a 3x3 matrix as follows: \(\mathbf{u} \times \mathbf{v} = \left| \begin{array} { c c c } i & j & k \ 0 & 1 & 0 \ 0 & 1 & 1 \end{array}\right|\). After calculating the determinant, you will get \(\mathbf{u} \times \mathbf{v} = -\mathbf{i}\).
03

Find the Magnitude of the Cross Product

To find the magnitude (length) of the cross product, take the square root of the sum of the squares of the components. In this case, the magnitude is: \(\| \mathbf{u} \times \mathbf{v} \|= \sqrt{(-1)^2} = 1\).
04

Determine the Area of the Parallelogram

Now, the area of the parallelogram formed by vectors \(\mathbf{u}\) and \(\mathbf{v}\) is the magnitude of their cross product. Thus, the area is 1.

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