Dot Product vs Cross Product: When to Use Which

Dot product and cross product look similar on paper, but they answer different questions. One gives a number that measures how much two vectors point the same way; the other gives a vector perpendicular to both whose length measures the area they span. This article will help you decide instantly which one a question needs, and use each for angles, projections, perpendicularity and areas.

Side-by-side summary

Why each one does its job

The dot product contains \(\cos\theta\), which is largest when the vectors point the same way and zero at \(90^\circ\). So it naturally measures alignment. Dividing by one length gives the projection, the “shadow” of one vector on another:

\[\text{projection of } \vec a \text{ on } \vec b = \frac{\vec a\cdot\vec b}{|\vec b|}.\]

The cross product contains \(\sin\theta\). Since \(|\vec a||\vec b|\sin\theta\) is the base times the height of the parallelogram formed by \(\vec a\) and \(\vec b\), the magnitude of the cross product is that area. Its direction, given by the right-hand rule, is perpendicular to the plane of \(\vec a\) and \(\vec b\).

Worked example 1: one pair of vectors, every quantity

Let \(\vec a = 2\hat i – \hat j + 2\hat k\) and \(\vec b = \hat i + 2\hat j + 2\hat k\).

Angle (dot product)

\(\vec a\cdot\vec b = 2(1) + (-1)(2) + 2(2) = 4\). Also \(|\vec a| = \sqrt{4+1+4} = 3\) and \(|\vec b| = \sqrt{1+4+4} = 3\). So

\[\cos\theta = \frac{4}{3\cdot 3} = \frac{4}{9},\qquad \theta = \cos^{-1}\frac{4}{9}\;(\approx 63.6^\circ).\]

Projection (dot product)

\[\text{projection of } \vec a \text{ on } \vec b = \frac{\vec a\cdot\vec b}{|\vec b|} = \frac{4}{3}.\]

Area and perpendicular vector (cross product)

\[\vec a\times\vec b = \begin{vmatrix}\hat i & \hat j & \hat k\\ 2 & -1 & 2\\ 1 & 2 & 2\end{vmatrix} = \hat i(-2-4) – \hat j(4-2) + \hat k(4+1) = -6\hat i – 2\hat j + 5\hat k.\]

\(|\vec a\times\vec b| = \sqrt{36 + 4 + 25} = \sqrt{65}\). So the parallelogram with adjacent sides \(\vec a, \vec b\) has area \(\sqrt{65}\) square units, and the triangle with these sides has area \(\frac{\sqrt{65}}{2}\).

A unit vector perpendicular to both \(\vec a\) and \(\vec b\) is \(\pm\dfrac{1}{\sqrt{65}}(-6\hat i – 2\hat j + 5\hat k)\). Check: \((-6)(2) + (-2)(-1) + 5(2) = 0\) and \((-6)(1) + (-2)(2) + 5(2) = 0\).

A built-in consistency check

Since \(\cos^2\theta + \sin^2\theta = 1\), we always have \((\vec a\cdot\vec b)^2 + |\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2\). Here \(16 + 65 = 81 = 9\times 9\). If this identity fails in your working, one of the products is wrong.

Worked example 2: area of a triangle from its vertices

Find the area of the triangle with vertices \(A(1,0,2)\), \(B(3,1,0)\) and \(C(2,-1,1)\).

Form two sides from the same vertex: \(\overrightarrow{AB} = 2\hat i + \hat j – 2\hat k\), \(\overrightarrow{AC} = \hat i – \hat j – \hat k\).

\[\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\hat i & \hat j & \hat k\\ 2 & 1 & -2\\ 1 & -1 & -1\end{vmatrix} = \hat i(-1-2) – \hat j(-2+2) + \hat k(-2-1) = -3\hat i – 3\hat k.\] \[\text{Area} = \frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right| = \frac{1}{2}\sqrt{9 + 9} = \frac{3\sqrt2}{2}\text{ sq units}.\]

The dot product would not help here: \(\overrightarrow{AB}\cdot\overrightarrow{AC} = 3\) gives the angle at \(A\), but you would still need a sine and two lengths. The cross product does it in one step.

Worked example 3: testing perpendicular and parallel

(a) Find \(\lambda\) so that \(\hat i + \lambda\hat j + 3\hat k\) is perpendicular to \(2\hat i – \hat j + \hat k\).

Perpendicular means dot product zero: \(2 – \lambda + 3 = 0\), so \(\lambda = 5\).

(b) Are \(2\hat i – \hat j + 3\hat k\) and \(-4\hat i + 2\hat j – 6\hat k\) parallel?

The second is \(-2\) times the first, so they are parallel (in opposite directions). Equivalently, their cross product is \(\vec 0\). For a quick test, compare ratios of components: \(\frac{-4}{2} = \frac{2}{-1} = \frac{-6}{3} = -2\).

Quick reference: which product for which question?

• Angle between two vectors → dot.
• Projection or component of \(\vec a\) along \(\vec b\) → dot, divided by \(|\vec b|\).
• Are two vectors perpendicular? → dot \(= 0\).
• Area of parallelogram with adjacent sides \(\vec a, \vec b\) → \(|\vec a\times\vec b|\).
• Area of a parallelogram from its diagonals \(\vec d_1, \vec d_2\) → \(\frac{1}{2}|\vec d_1\times\vec d_2|\).
• Area of a triangle → \(\frac{1}{2}|\text{side}\times\text{side}|\).
• Vector perpendicular to two vectors → cross; divide by its magnitude for a unit vector.
• Are two vectors parallel? → cross \(= \vec 0\), or proportional components.

Common mistakes

• Sign error in the middle term of the cross product determinant: it is \(-\hat j(a_1b_3 – a_3b_1)\).
• Dividing by the wrong magnitude for projection. Projection of \(\vec a\) on \(\vec b\) divides by \(|\vec b|\).
• Forgetting the \(\frac{1}{2}\) for a triangle, or adding it for a parallelogram with adjacent sides.
• Forgetting \(\pm\) when asked for a unit vector perpendicular to two vectors; there are two directions.
• Treating the cross product as a number. It is a vector; only its magnitude is an area.

Practise next

Strengthen each idea separately on the scalar product and vector product topic pages, then try mixed questions on the Vector Algebra chapter page. The cross product reappears in shortest distance between lines, so it is worth mastering now. Create a free student account for timed practice tests.

Frequently asked questions