So far, addition, subtraction, and scalar multiplication of vectors have been introduced and explored.

While each of these operations has an intuitive geometrical interpretation, they alone are unable to handle certain applications of vectors.

In this Module

• We will develop two new operations that will allow us to extend our ability to apply vectors to physical and geometrical situations.
• We will examine a new vector operation: the dot product.

Definition

Find the dot product of \(\vec{u}\) and \(\vec{v}\) given that \(\lvert \vec{u} \rvert =8, \lvert \vec{v} \rvert =3 \), and the angle between \(\vec{u}\) and \(\vec{v}\) is

a. \(\theta=40^\circ\)

b. \(\theta = \dfrac{3\pi}{4}\)

Solution

a.

b.

Note:

Since \(\lvert \vec{u}\rvert, \lvert \vec{v}\rvert\), and \(\cos(\theta)\) are all scalars (real numbers), their product will be a scalar.

Thus, unlike the previously learned vector operations (which produced vectors as answers), the dot product always yields a scalar answer.

For this reason, dot product is often referred to as the scalar product.

Prove that if the dot product of two non-zero vectors, \(\vec{u}\) and \(\vec{v}\), is equal to zero, then \(\vec{u}\) and \(\vec{v}\) must be perpendicular.

Solution

We are required to prove \( \vec{u} \perp \vec{v}\) given that \( \vec{u}\cdot \vec{v}=0\).

Proof

Since \( \lvert \vec{u}\rvert \neq 0 \) and \( \lvert \vec{v} \rvert \neq 0 \), it must be that \( \cos(\theta) = 0\).

Solving this equation, we get \(\theta = \pm 90^\circ\).

Therefore, \(\vec{u}\) is perpendicular to \(\vec{v}\).

Is the converse true? That is, if non-zero vectors, \(\vec{u}\) and \(\vec{v}\), are perpendicular, then does \(\vec{u}\cdot\vec{v}=0\)?

This example proves an important property of the dot product of two vectors:

 Two vectors are orthogonal if they are perpendicular, but includes the case where either (possibly both) vectors are \(\vec{0}\).