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Pure · Vectors

Chapter 1 · 4

The idea

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.

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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

Three facts in one number

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

This 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.)

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