## Scalar/Dot Product

Given 3 dimensional vectors we define their dot product as follows;

If $$v=ai+bj+ck$$ and $$w=di+ej+fk$$ then
$$v.w=(ai+bj+ck ).(di+ej+fk)$$
$$v.w=ad+be+cf$$

LESSON 1

If $$t=-i+3j-2k$$ and $$r=2i-4k$$ determine $$r.t$$

SOLUTION
$$r.t=(-i+3j-2k).(2i-4k)$$
$$r.t=(-1)(2)+(3)(0)+(-2)(-4)$$
$$r.t=6$$

TRY THIS

Complete Questions 2 and 3 from worksheet.

## ANGLE BETWEEN 2 VECTORS

We can use the scalar product to find the angle between two vectors, thanks to the following formula:
$$a.b=|a||b| \cos⁡θ$$
where $$θ$$ is the angle between the vectors.

Notice that the vectors $$a$$ and $$b$$ are going away from the angle, $$θ$$.

An important fact is that two vectors are perpendicular (orthogonal) if and only if their dot product is zero. This is because if $$θ=90°$$, then $$a.b=0$$ (Recall: $$\cos⁡ 90°=0$$)

LESSON 1

Given that $$a=\begin{pmatrix} 2\\-2 \\1 \end{pmatrix}$$, $$b=\begin{pmatrix} 2\\6 \\3 \end{pmatrix}$$ and $$c=\begin{pmatrix} p\\p \\p+1 \end{pmatrix}$$, find

(i) The angle between the directions of $$a$$ and $$b$$.
(ii) The value of $$p$$ for which $$b$$ and $$c$$ are perpendicular
SOLUTION

Part (i)

$$a.b=|a||b| \cos⁡θ$$
$$\displaystyle{\begin{pmatrix} 2\\-2 \\1 \end{pmatrix}.\begin{pmatrix} 2\\6 \\3 \end{pmatrix}\over \sqrt{2^2+(-2)^2+1}\sqrt{2^2+6^2+3^2 }}=\cos⁡θ$$
$$\displaystyle{(2)(2)+(-2)(6)+(1)(3)\over \sqrt{9}\sqrt{49}}=\cos⁡θ$$
$$\displaystyle-{5\over21}=\cos⁡θ$$
$$\displaystyle θ=\cos^{-1}⁡(-{5\over21})=103.8°$$

Part (ii)

If $$b$$ and $$c$$ are perpendicular then $$b.c =0$$
$$\begin{pmatrix} 2\\6 \\3 \end{pmatrix}.\begin{pmatrix} p\\p \\p+1 \end{pmatrix}=0$$
$$2p+6p+3(p+1)=0$$
$$2p+6p+3p+3=0$$
$$11p=-3$$
$$p=-{3\over11}$$

TRY THIS

Complete Questions 4 - 7 from worksheet.