Proving Pythagoras theorem with perpendicular vectors

Key concept: This proof uses vector dot products to connect perpendicularity with squared side lengths.

Step by step

Key idea

If two vectors are perpendicular, their dot product is zero. That single fact is enough to build a clean vector proof of Pythagoras' theorem.

Let the two perpendicular vectors be represented by $\mathbf{b}$ and $\mathbf{c}$, so $\mathbf{b}\cdot\mathbf{c}=0$. If the diagonal or resultant vector is $\mathbf{a}=\mathbf{b}+\mathbf{c}$, then the square of its length is

$|\mathbf{a}|^2=(\mathbf{b}+\mathbf{c})\cdot(\mathbf{b}+\mathbf{c}).$

Vector expansion

Expand the dot product using distributive properties:

=\mathbf{b}\cdot\mathbf{b}+2\mathbf{b}\cdot\mathbf{c}+\mathbf{c}\cdot\mathbf{c}.$$ Because the vectors are perpendicular, $\mathbf{b}\cdot\mathbf{c}=0$. So this becomes $$|\mathbf{a}|^2=|\mathbf{b}|^2+|\mathbf{c}|^2.$$ That is exactly the Pythagorean relationship. In geometric language, the vector $\mathbf{a}$ plays the role of the hypotenuse, while $\mathbf{b}$ and $\mathbf{c}$ are the two perpendicular sides. ## Why this proves Pythagoras' theorem The theorem says that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. The vector proof does not need a diagram-specific calculation; it relies on a general algebraic identity. If you want a formal exam-style ending, state: Since $\mathbf{b}\perp\mathbf{c}$, we have $\mathbf{b}\cdot\mathbf{c}=0$. Hence $$|\mathbf{a}|^2=|\mathbf{b}+\mathbf{c}|^2=(\mathbf{b}+\mathbf{c})\cdot(\mathbf{b}+\mathbf{c})=|\mathbf{b}|^2+|\mathbf{c}|^2.$$ Therefore, Pythagoras' theorem is proved. ## Common exam pitfall Do not write $|\mathbf{b}+\mathbf{c}|=|\mathbf{b}|+|\mathbf{c}| $; lengths do not add like that in general. The correct move is to square the resultant and expand the dot product. Also, you must explicitly use perpendicularity to eliminate the cross term. Without the step $\mathbf{b}\cdot\mathbf{c}=0$, the argument is incomplete. ### Pitfall alert A common mistake is to treat vectors like ordinary lengths and write the diagonal length as the sum of the two side lengths. That is false except in very special cases. Another frequent error is expanding $(\mathbf{b}+\mathbf{c})^2$ without using dot products, which hides the reason the cross term disappears. In vector proofs, the zero dot product from perpendicular vectors is the key step, so it should be stated clearly and used before the final equality. ### Try different conditions If the question changes to "show that a triangle with sides represented by perpendicular vectors satisfies Pythagoras", the same method still works. For example, if one side is $\mathbf{u}$, the other is $\mathbf{v}$, and the hypotenuse is $\mathbf{u}+\mathbf{v}$, then $|\mathbf{u}+\mathbf{v}|^2=|\mathbf{u}|^2+|\mathbf{v}|^2$ whenever $\mathbf{u}\cdot\mathbf{v}=0$. If the vectors are given by coordinates, you can also prove perpendicularity by checking their dot product directly first, then apply the same expansion. ### Further reading dot product, perpendicular vectors, Pythagorean theorem