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Pure · Vectors

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The scalar (dot) product

The scalar product a·b = |a||b|cos θ and its component form — using it to find the angle between two vectors, to test for perpendicularity, and the rule that the angle at a vertex needs both vectors pointing away from that vertex.

Pure · Vectors

The scalar (dot) product

The scalar product a·b = |a||b|cos θ and its component form — using it to find the angle between two vectors, to test for perpendicularity, and the rule that the angle at a vertex needs both vectors pointing away from that vertex.

Why it works

Two vectors have a length each and an angle between them. The scalar product packages all three into one number: a⋅b=∣a∣ ∣b∣cos⁡θ,\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta, where θ\theta is the angle between the two vectors. The result is a scalar — an ordinary number, not a vector. That is what the name is telling you, and why writing a vector as the answer is always wrong.

The component form is what you actually compute with. For a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} and b=b1i+b2j+b3k\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}, a⋅b=a1b1+a2b2+a3b3.\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3. Multiply matching components, then add — one number out. (It works because i⋅i=j⋅j=k⋅k=1\mathbf{i}\cdot\mathbf{i} = \mathbf{j}\cdot\mathbf{j} = \mathbf{k}\cdot\mathbf{k} = 1 and any two different unit vectors are perpendicular, so their products vanish.)

Finding an angle. Equating the two forms and rearranging, cos⁡θ=a⋅b∣a∣ ∣b∣.\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}. Compute the dot product on top and the two magnitudes underneath, then take cos⁡−1\cos^{-1}. A negative dot product is perfectly normal — it simply means the angle is obtuse, and cos⁡−1\cos^{-1} will return an answer between 90∘90^\circ and 180∘180^\circ. Never discard the minus sign to "tidy" the answer.

The perpendicularity test. If θ=90∘\theta = 90^\circ then cos⁡θ=0\cos\theta = 0, so for non-zero vectors a⊥b  ⟺  a⋅b=0.\mathbf{a} \perp \mathbf{b} \iff \mathbf{a} \cdot \mathbf{b} = 0. This is the workhorse: it turns a geometric condition into one linear equation, which is how questions ask you to find an unknown component.

Useful facts. a⋅b=b⋅a\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} (order does not matter), and a⋅a=∣a∣2\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 (since θ=0\theta = 0).

The angle at a vertex — the trap. To find angle ABCABC, the angle at BB, you need the two vectors that point away from BB: BA→\overrightarrow{BA} and BC→\overrightarrow{BC}. Using AB→\overrightarrow{AB} instead reverses one vector, which flips the sign of the dot product and hands you 180∘−θ180^\circ - \theta — the wrong angle, and it will look plausible. Always start both vectors at the vertex named in the middle.