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Chapter 5: Problem 27

Find the area of the parallelogram that has the vectors as adjacent sides. $$\mathbf{u}=(3,2,-1), \quad \mathbf{v}=(1,2,3)$$

Short Answer

Expert verified
The area of the parallelogram is \(\sqrt{96}\)

Step by step solution

01

Calculate the cross product

The first step is to calculate the cross product of the given vectors \(\mathbf{u}\) and \(\mathbf{v}\). The cross product of \(\mathbf{u}=(u_1, u_2, u_3)\) and \(\mathbf{v}=(v_1, v_2, v_3)\) is given by \(\mathbf{u} \times \mathbf{v}= ((u_2v_3 - u_3v_2), (u_3v_1 - u_1v_3), (u_1v_2 - u_2v_1))\). So in this case \(\mathbf{u}=(3,2,-1)\) and \(\mathbf{v}=(1,2,3)\) hence \(\mathbf{u} \times \mathbf{v} = ((2x3 - (-1)x2), ((-1)x1 - 3x1), (3x2 - 2x1)) = (6 + 2, -1 - 3, 6 - 2) = (8, -4, 4)\)
02

Calculate the magnitude of the cross product

Once the cross product \(\mathbf{u} \times \mathbf{v}\) has been calculated, the next step is to find its magnitude, which will be the area of the parallelogram formed by \(\mathbf{u}\) and \(\mathbf{v}\). The magnitude of a vector \(\mathbf{w}=(w_1,w_2,w_3)\) is given by \(\sqrt{w_1^2 + w_2^2 + w_3^2}\). Hence the area of the parallelogram \(= ||\mathbf{u} \times \mathbf{v}|| = \sqrt{8^2 + (-4)^2 + 4^2} = \sqrt{64 + 16 + 16} = \sqrt{96}\)

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Most popular questions from this chapter

Find the cross product of the unit vectors [where \(\mathbf{i}=(1,0,0), \mathbf{j}=(0,1,0), \text { and } \mathbf{k}=(0,0,1)] .\) Sketch your result. $$\mathbf{j} \times \mathbf{i}$$

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Find the area of the parallelogram that has the vectors as adjacent sides. $$\mathbf{u}=(2,-1,0), \quad \mathbf{v}=(-1,2,0)$$
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