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Vectors · Topic 7 of 11

Scalar/Dot Product (1)

Video lesson · from 36:282 worked examples

One lesson video covers all of Vectors, so it opens at 36:28 for this topic — not from the beginning.

Theory

The dot product can be thought of as directional multiplication. Multiplying two vectors means we are applying the directional growth of one vector to another. The result is how much stronger we have made the original vector.

The final result of the dot product process can be:

  • Zero – no growth in the original direction
  • Positive – some growth in the original direction
  • Negative – negative/reverse growth in the original direction

The dot product (scalar product), denoted a⃗⋅b⃗\vec{a} \cdot \vec{b}, can be calculated as follows (Rectangular perspective):

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3

where a⃗=(a1a2a3)\vec{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} and b⃗=(b1b2b3)\vec{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}

⚠️ Common Examiner Traps

  • The result is a number, not a vector: the scalar product produces a scalar. An answer written as a column vector shows the method was misunderstood.
  • Multiply matching components, then add: a1b1+a2b2+a3b3a_1b_1 + a_2b_2 + a_3b_3 — it is not component-by-component multiplication kept as a vector.
  • Mind the negatives: a single sign error changes the answer completely, and there is no obvious check.
  • Zero means perpendicular: if the scalar product is zero, the vectors are at right angles — a fact many questions depend on.

Worked examples

Example 1

Find p⃗⋅q⃗\vec{p} \cdot \vec{q}, given that p⃗=(12−3)\vec{p} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} and q⃗=(223)\vec{q} = \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}.

p⃗⋅q⃗=(1)(2)+(2)(2)+(−3)(3)=2+4−9=−3\vec{p} \cdot \vec{q} = (1)(2) + (2)(2) + (-3)(3) = 2 + 4 - 9 = -3

Example 2

If A is the point (2,3,9)(2, 3, 9), B(1,4,−2)(1, 4, -2) and C(−1,3,−6)(-1, 3, -6), calculate AB⃗⋅AC⃗\vec{AB} \cdot \vec{AC}.

First find AB⃗\vec{AB} and AC⃗\vec{AC}:

AB⃗=b⃗−a⃗=(14−2)−(239)=(−11−11)\vec{AB} = \vec{b} - \vec{a} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ 9 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ -11 \end{pmatrix}
AC⃗=c⃗−a⃗=(−13−6)−(239)=(−30−15)\vec{AC} = \vec{c} - \vec{a} = \begin{pmatrix} -1 \\ 3 \\ -6 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ 9 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ -15 \end{pmatrix}

Calculate the dot product:

AB⃗⋅AC⃗=(−1)(−3)+(1)(0)+(−11)(−15)=3+0+165=168\vec{AB} \cdot \vec{AC} = (-1)(-3) + (1)(0) + (-11)(-15) = 3 + 0 + 165 = 168