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Chapter 5: Problem 34

Find \(\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) .\) This quantity is called the triple scalar product of \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\). $$\mathbf{u}=(2,0,1), \quad \mathbf{v}=(0,3,0), \quad \mathbf{w}=(0,0,1)$$

Short Answer

Expert verified
The triple scalar product of \(\mathbf{u}, \mathbf{v}\), and \(\mathbf{w}\) is 6.

Step by step solution

01

Calculate Cross product of vectors v and w

We start by calculating the cross product of \(\mathbf{v}\) and \(\mathbf{w}\), which for 3D vectors is defined as \( \mathbf{v} \times \mathbf{w} = (v_{2}w_{3} - v_{3}w_{2}, v_{3}w_{1} - v_{1}w_{3}, v_{1}w_{2} - v_{2}w_{1} ) = (3*1 - 0*0, 0*0 - 0*1, 0*1 - 3*0) = (3, 0, 0)\)
02

Calculate the dot product

Now, we perform the dot product of the result from step 1, \( \mathbf{v} \times \mathbf{w}\), and the vector \(\mathbf{u}\). For 3D vectors, the dot product is given by \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) = u_{1}(v_{2}w_{3} - v_{3}w_{2}) + u_{2}(v_{3}w_{1} - v_{1}w_{3}) + u_{3}(v_{1}w_{2} - v_{2}w_{1}) = 2*3 + 0*0 + 1*0 = 6\)
03

Conclusion

The triple scalar product of \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) is 6.

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Most popular questions from this chapter

Show that the volume of a parallelepiped having \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) as adjacent sides is the triple scalar product \(|\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})|\).

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. $$\mathbf{j} \times \mathbf{k}$$

(a) find the linear least squares approximating function \(g\) for the function \(f\) and \((b)\) use a graphing utility to graph \(f\) and \(g\). $$f(x)=\sin x,-\pi / 2 \leq x \leq \pi / 2$$

Use a graphing utility with vector capabilities to find \(\mathbf{u} \times \mathbf{v}\) and then show that it is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). $$\mathbf{u}=(1,2,-1), \quad \mathbf{v}=(2,1,2)$$

Prove that \(\mathbf{u} \times \mathbf{0}=\mathbf{0} \times \mathbf{u}=\mathbf{0}\).
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