Leave lesson

Pure · Vectors

1 / 11

Vectors in three dimensions

Extending vectors to 3D — the i, j, k and column forms, magnitude by Pythagoras in three dimensions, the distance between two points, unit vectors, and the test for two vectors being parallel.

Pure · Vectors

Vectors in three dimensions

Extending vectors to 3D — the i, j, k and column forms, magnitude by Pythagoras in three dimensions, the distance between two points, unit vectors, and the test for two vectors being parallel.

Why it works

A point in space needs three numbers, so we add a third axis zz perpendicular to both xx and yy. Every rule you already know for 2D vectors carries over unchanged — you just carry one more component: a=a1i+a2j+a3k=(a1a2a3).\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}. Addition, subtraction and multiplication by a scalar are still done component by component. Nothing new is being defined; the dimension is the only change.

Magnitude is Pythagoras used twice. For a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}, ∣a∣=a1 2+a2 2+a3 2.|\mathbf{a}| = \sqrt{a_1^{\,2} + a_2^{\,2} + a_3^{\,2}}. The reason: a12+a22\sqrt{a_1^2 + a_2^2} is the length of the shadow in the xyxy-plane, and the zz-component is perpendicular to that shadow, so a second Pythagoras gives (a12+a22)2+a32\sqrt{\left(\sqrt{a_1^2+a_2^2}\right)^2 + a_3^2}, which is the formula above. Squaring kills the signs, so negative components do not make a length negative.

Distance between two points. If AA and BB have position vectors a\mathbf{a} and b\mathbf{b}, then AB→=b−a,\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, and the distance AB=∣b−a∣AB = |\mathbf{b} - \mathbf{a}|. The order matters for the vector (going the other way reverses it, BA→=−AB→\overrightarrow{BA} = -\overrightarrow{AB}) but not for the distance, because the squaring removes the sign. A reliable way to remember it: to get from AA to BB you go back along a\mathbf{a} and out along b\mathbf{b}, so it is finish minus start.

Unit vectors. A unit vector has magnitude 11. To build one in the direction of a\mathbf{a}, divide every component by ∣a∣|\mathbf{a}|: a^=1∣a∣ a.\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|}\,\mathbf{a}. Dividing a vector by a positive scalar keeps its direction and scales its length, so ∣a^∣=1|\hat{\mathbf{a}}| = 1 exactly.

Parallel vectors. Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, b=λa\mathbf{b} = \lambda\mathbf{a}. That means the components are proportional, not equal — and the same λ\lambda must work for all three components. A negative λ\lambda still counts as parallel; the vectors simply point in opposite directions.