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Chapter 11: Problem 34

Show that \(\vec{A} \cdot(\vec{A} \times \vec{B})=0\) for any vectors \(\vec{A}\) and \(\vec{B}\).

Short Answer

Expert verified
The dot product of vector A and the cross product of vectors A and B equals zero because the result of a cross product is a vector that is orthogonal to the initial vectors, and the dot product of orthogonal vectors is zero due to the cosine of 90 degrees yielding zero.

Step by step solution

01

Understand the properties of dot and cross products

Recall that the dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{A} \cdot \vec{B} = |A||B| \cos(\theta)\), where \(\theta\) is the angle in-between them. If the vectors are perpendicular, then \(\theta = 90^{\circ}\) and \(\cos(\theta)=0\). The cross product \(\vec{A} \times \vec{B}\) yields a vector that is orthogonal to the plane formed by \(\vec{A}\) and \(\vec{B}\).
02

Apply these properties to the given equation

We have \(\vec{A} \cdot (\vec{A} \times \vec{B})\). Consider vector \(\vec{A}\) and the result of the cross product, \(\vec{A} \times \vec{B}\). These vectors are perpendicular to each other according to the properties of the cross product. So, if we take the dot product of these two vectors, the result will be zero because \(\cos(\theta) = 0\) for \(\theta = 90^{\circ}\). That is, \(\vec{A} \cdot (\vec{A} \times \vec{B}) = |A||A \times B| \cos(90^{\circ}) = 0\).

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