13.04 Vectors in 3D

Ideas about vectors in the plane extend easily to vectors in three-dimensional space.

We can represent them as arrows with the length of the arrow corresponding to the magnitude of the vector.  We can move vectors parallel with themselves so that the tail of the arrow is at the origin. The direction of the arrow is then determined by the position of the arrowhead in the three-dimensional coordinate system.

For example, an arrowhead with coordinates $\left(2,3,-1\right)$(2,3,−1) represents a vector pointing $2$2 units in the $x$x-direction, $3$3 units in the $y$y-direction and $-1$−1 unit in the $z$z-direction. Its length is found by Pythagoras's theorem to be $\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$√22+32+(−1)2=√14.

The mutually perpendicular unit vectors notated $\mathbf{i}$i, $\mathbf{j}$j and $\mathbf{k}$k are aligned with the coordinate axes $x$x, $y$y and $z$z respectively so that a vector in coordinate form, like $\left(2,3,-1\right)$(2,3,−1), is equivalent to $2\mathbf{i}+3\mathbf{j}-\mathbf{k}$2i+3j−k in terms of the orthonormal basis vectors.

 

Magnitude of vector in 3D

We calculate the length of a three-dimensional vector by applying Pythagoras's theorem, just as is done for two-dimensional vectors. Thus,

$|\mathbf{v}|=|(a,b,c)|=\sqrt{a^2+b^2+c^2}$|v|=|(a,b,c)|=√a2+b2+c2.

Distance between two vectors

The distance between two vectors $\mathbf{u}=(u_1,u_2,u_3)$u=(u1​,u2​,u3​)  and $\mathbf{v}=(v_1,v_2,v_3)$v=(v1​,v2​,v3​) is given by

$|\mathbf{v}-\mathbf{u}|=\sqrt{\left(v_1-u_1\right)^2+\left(v_2-u_2\right)^2+\left(v_3-u_3\right)^2}$|v−u|=√(v1​−u1​)2+(v2​−u2​)2+(v3​−u3​)2.

 

Position vectors

Given points or position vectors $P(a,b,c)$P(a,b,c) and $Q(p,q,r)$Q(p,q,r), the vector representing the line segment joining $P$P and $Q$Q is given by

$\vec{PQ}=(p-a,q-b,r-c)$→PQ=(p−a,q−b,r−c).

 

Worked example

example 1

Find the position vector of $\overrightarrow{PQ}$›‹PQ if $P=(0,-3,4)$P=(0,−3,4) and $Q=(2,5,3)$Q=(2,5,3). Give your answer in the form $a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$ai+bj+ck.

Think: We find the position vector by finding the difference between the $x$x, $y$y and $z$z coordinates, starting with the coordinates of point $Q$Q and subtracting those from point $P$P.

Do: $\overrightarrow{PQ}=(2-0,5--3,3-4)$›‹PQ=(2−0,5−−3,3−4) 

  So    $\overrightarrow{PQ}=(2,8,-1)$›‹PQ=(2,8,−1) 

   which is  $\overrightarrow{PQ}=2\mathbf{i}+8\mathbf{j}-\mathbf{k}$›‹PQ=2i+8j−k.

Equation of a plane

A plane passing through the origin is represented by the Cartesian equation $ax+by+cz=0$ax+by+cz=0. where $a$a, $b$b and $c$c are constants. A particular example might be $x-3y+2z=0$x−3y+2z=0.  A vector lying in this plane is given by $a\mathbf{i}+b\mathbf{j}+\frac{a-3b}{-2}\mathbf{k}$ai+bj+a−3b−2​k. 

Equation of a sphere

We specify a sphere with radius $r$r and centre $(a,b,c)$(a,b,c) by means of the Cartesian formula $r^2=(x-a)^2+(y-b)^2+(z-c)^2$r2=(x−a)2+(y−b)2+(z−c)2.

A corresponding vector formulation is 

$\left|(x-a)\mathbf{i}+(y-b)\mathbf{j}+(z-c)\mathbf{k}\right|=r$|(x−a)i+(y−b)j+(z−c)k|=r

This says that the distance from a point $(x,y,z)$(x,y,z) on the sphere to the centre point $(a,b,c)$(a,b,c) is a constant $r$r.

Practice questions

Question 1

Question 2

Question 3

Question 4