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

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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 \(b.c =0\)
\(\begin{pmatrix} 2\\6 \\3 \end{pmatrix}.\begin{pmatrix} p\\p \\p+1 \end{pmatrix}=0\)
\(2p+6p+3(p+1)=0\)
\(2p+6p+3p+3=0\)
\(11p=-3\)
\(p=-{3\over11}\)

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Complete Questions 4 - 7 from worksheet.