Question

Let "a" and "b" be two vectors of unit length, and let "θ" be the angle between them. Show that:

cos(θ)=a⋅b

where ⋅ denotes the dot product.

Answer 1

To show that cos(θ) = a·b for unit vectors a and b, simply apply the dot product formula, a·b = |a||b|cos(θ), and consider that the magnitudes of unit vectors are 1, resulting in a·b = cos(θ).

Step-by-step explanation:

The student asked to show that for two vectors of unit length a and b, with the angle θ between them, the dot product a·b is equal to cos(θ). This is a fundamental concept in vector mathematics. The dot product, also known as the scalar product, can be expressed as a·b = |a||b|cos(θ), where |a| and |b| are the magnitudes of vectors a and b, respectively, and θ is the angle between them. Since both vectors have unit length, their magnitudes are 1, simplifying the expression to a·b = cos(θ).

In mathematical terms, the dot product of two vectors a and b with respective components Ax, Ay, Az and Bx, By, Bz is given as a·b = Ax Bx + Ay By + Az Bz. When these vectors are of unit length, their dot product equals the cosine of the angle between them. Therefore, this calculation directly yields the cosine of the angle when the vector magnitudes are 1.

When vectors are perpendicular, the angle between them is 90 degrees and the dot product is zero, reflecting that cos(90°) = 0. Similarly, for parallel vectors, the angle is 0 degrees and the dot product equals 1, as cos(0°) = 1.