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What is the dot product of two orthogonal vectors?

Short Answer

Expert verified
Answer: The dot product of two orthogonal vectors is 0.

Step by step solution

01

Define dot product

The dot product of two vectors, A = and B = , is defined as follows: Dot product(A, B) = a1 * b1 + a2 * b2 + a3 * b3
02

Define orthogonal vectors

Two vectors are orthogonal if they are perpendicular to each other, which means the angle between the two vectors is 90 degrees. We can use the dot product to check if two vectors are orthogonal by the following property: If A and B are orthogonal vectors, then Dot(A, B) = 0. This property arises from the fact that the dot product equals to the product of the magnitudes of the vectors multiplied by the cosine of the angle between the vectors. When the angle between the vectors is 90 degrees, then cosine(90) = 0.
03

Calculate the dot product of orthogonal vectors

We know that the dot product of orthogonal vectors equals zero, due to the property discussed in Step 2. Therefore: Dot product(orthogonal vectors) = 0

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Most popular questions from this chapter

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Assuming the curve lies in a plane, show that it is a circle centered at the origin with radius \(R\) provided \(a^{2}+c^{2}+e^{2}=b^{2}+d^{2}+f^{2}=R^{2}\) and \(a b+c d+e f=0\).

For the given points \(P, Q,\) and \(R,\) find the approximate measurements of the angles of \(\triangle P Q R\). $$P(1,-4), Q(2,7), R(-2,2)$$

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Prove that for integers \(m\) and \(n\), the curve $$\mathbf{r}(t)=\langle a \sin m t \cos n t, b \sin m t \sin n t, c \cos m t\rangle$$ lies on the surface of a sphere provided \(a^{2}+b^{2}=c^{2}\).

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