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Vectors in 3D

The geometric meaning of a vector

Think of a vector as an arrow. The arrow has two important features:

  1. Direction: which way it points.
  2. Magnitude: how long it is.

So a vector is really a piece of geometric information about a move from one place to another place. If you start at some point and walk along the arrow, the vector tells you both the direction of the walk and the distance you travel.

What are the two main pieces of information carried by a vector?

Answer:

A vector describes a direction and a magnitude.

Vectors in 2D and 3D

In 2D, a vector looks like this:

A vector in 2D

Vector uāƒ— =(,)
šŸ’” Tip: Drag the arrow head to change direction/magnitude, drag the circular tail to move the starting point, or drag the line to translate the entire vector.
uāƒ—(3.0, 2.0)

In 3D, the same idea becomes an ordered triple

vāƒ—=(x,y,z).\vec{v} = (x, y, z).

Geometrically, we can think of it as the arrow from the origin to the point (x,y,z)(x,y,z).

A vector in 3D space

Vector endpoint: (3, 2, 4)
Magnitude: 5.39
šŸ’” Tip: in vector mode, drag inside the graph to move the endpoint. Hold Shift while dragging to adjust z.

What changes when we move from a 2D vector (x,y)(x,y) to a 3D vector (x,y,z)(x,y,z)?

Answer:

We add one more coordinate, so the vector can describe movement in one more independent direction.

Magnitude of a 3D vector

The magnitude of vāƒ—=(x,y,z)\vec{v} = (x,y,z) is its length:

āˆ„vāƒ—āˆ„=x2+y2+z2.\|\vec{v}\| = \sqrt{x^2 + y^2 + z^2}.

This is the 3D version of the Pythagorean theorem.

For example, if

vāƒ—=(2,āˆ’1,6),\vec{v} = (2, -1, 6),

then

āˆ„vāƒ—āˆ„=22+(āˆ’1)2+62=41.\|\vec{v}\| = \sqrt{2^2 + (-1)^2 + 6^2} = \sqrt{41}.

Compute the magnitude of

uāƒ—=(āˆ’3,4,12).\vec{u} = (-3, 4, 12).

Answer:

āˆ„uāƒ—āˆ„=(āˆ’3)2+42+122=9+16+144=169=13.\|\vec{u}\| = \sqrt{(-3)^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13.

Let

wāƒ—=(1,2,2).\vec{w} = (1, 2, 2).

Find āˆ„wāƒ—āˆ„\|\vec{w}\|.

Answer:

āˆ„wāƒ—āˆ„=12+22+22=9=3.\|\vec{w}\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3.

Standard basis vectors in 3D

The standard basis vectors in 3D are

i=(1,0,0),j=(0,1,0),k=(0,0,1).\mathbf{i} = (1,0,0), \quad \mathbf{j} = (0,1,0), \quad \mathbf{k} = (0,0,1).

They point along the positive xx-, yy-, and zz-axes respectively.

Any vector vāƒ—=(x,y,z)\vec{v} = (x,y,z) can be read as moving xx units in the i\mathbf{i} direction, yy units in the j\mathbf{j} direction, and zz units in the k\mathbf{k} direction.

For example,

(3,āˆ’2,5)(3,-2,5)

means 3 units in the xx direction, 2 units backward in the yy direction, and 5 units upward in the zz direction.

Which basis vector points along the positive zz-axis?

Answer:

k=(0,0,1).\mathbf{k} = (0,0,1).

What do the basis vectors i\mathbf{i} and j\mathbf{j} represent?

Answer:

i=(1,0,0),j=(0,1,0).\mathbf{i} = (1,0,0), \quad \mathbf{j} = (0,1,0).

So i\mathbf{i} points along the positive xx-axis and j\mathbf{j} points along the positive yy-axis.

Interpret the vector

(āˆ’2,5,1)(-2, 5, 1)

in terms of the coordinate directions.

Answer:

It means 2 units in the negative xx direction, 5 units in the positive yy direction, and 1 unit in the positive zz direction.

Unit vectors

If vāƒ—\vec{v} is a nonzero vector, then the unit vector in the direction of vāƒ—\vec{v} is

vāƒ—^=vāƒ—āˆ„vāƒ—āˆ„.\hat{\vec{v}} = \frac{\vec{v}}{\|\vec{v}\|}.

It has length 1, but it points in the same direction as vāƒ—\vec{v}.

For example, if

vāƒ—=(2,4,4),\vec{v} = (2, 4, 4),

then

āˆ„vāƒ—āˆ„=22+42+42=36=6,\|\vec{v}\| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{36} = 6,

so the unit vector is

vāƒ—^=(26,46,46)=(13,23,23).\hat{\vec{v}} = \left(\frac{2}{6}, \frac{4}{6}, \frac{4}{6}\right) = \left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right).

Find the unit vector in the direction of

uāƒ—=(3,0,4).\vec{u} = (3, 0, 4).

Answer:

āˆ„uāƒ—āˆ„=32+02+42=5,\|\vec{u}\| = \sqrt{3^2 + 0^2 + 4^2} = 5,

so

uāƒ—^=(35,0,45).\hat{\vec{u}} = \left(\frac{3}{5}, 0, \frac{4}{5}\right).

Is the vector

(12,12,12)\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}}\right)

a unit vector?

Answer:

(12)2+(12)2+(12)2=14+14+12=1.\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1.

Yes, it is a unit vector.

Normalize the vector

wāƒ—=(āˆ’6,2,3).\vec{w} = (-6, 2, 3).

Answer:

First compute its magnitude:

āˆ„wāƒ—āˆ„=(āˆ’6)2+22+32=49=7.\|\vec{w}\| = \sqrt{(-6)^2 + 2^2 + 3^2} = \sqrt{49} = 7.

Then the unit vector is

wāƒ—^=(āˆ’67,27,37).\hat{\vec{w}} = \left(-\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right).