Examples for Cartesian coordinates in three dimensions

Example 1

Find the magnitude of the vector x = 4i - 2j + 3k, the unit vector in the direction of x and the vector of magnitude 5 in the direction opposite to x.

Solution

We have |x| = V~ -2--------2---2- 4 + (- 2) + 3

Therefore the unit vector in the direction of x is the vector

1 V~ ---(4i - 2j + 3k) 29

and the vector of magnitude 5 in the direction opposite to x is

V~ 1-(- 20i + 10j - 15k) 29

Example 2

Given the vectors u = 3i - j + 2k and v = 2i + 3j - k, find Cartesian forms of the vectors u + v, u - v, 2u + 3v.

Solution

First,

u + v = (3i- j + 2k) + (2i + 3j - k) = 5i + 2j + k.

Next,

u - v = (3i- j + 2k) - (2i + 3j - k) = i- 4j + 3k.

and finally,

2u + 3v = 2(3i- j + 2k) + 3(2i + 3j- k) = 12i + 7j + k.

Example 3

Given that P₁(-1, 2, 3) and P₂(3, 3, 4) are two points in space, find the Cartesian form of the vector -P-1-->P2 .

Solution

-- --> P1P2 = (3- (-1))i + (3- 2)j + (4 - 3)k = 4i + j + k

Example 4

Show that the vectors

(3 + 4s)i + (4 - 3s)j + 5tk

and

(3s + 4t)i + 5j + (4s - 3t)k

have the same magnitude.

Solution

Both vectors are in Cartesian form and their lengths can be calculated using the formula

V~ ------------ |xi + yj + zk|= x2 + y2 + z2.

We have

V~ ---------------------------- |(3 + 4s)i + (4 - 3s)j + 5tk |= (3 + 4s)2 + (4 - 3s)2 + (5t)2 V~ -------------------------------------- = 9 + 24s + 16s2 + 16 - 24s + 9s2 + 25t2 V~ ---------2-----2 = 25 + 25s + 25t

and

V~ --------------------------- |(3s + 4t)i + 5j + (4s - 3t)k |= (3s + 4t)2 + 52 + (4s - 3t)2 V~ --2------2------------------2----2-------- = V~ 9s-+-16t-+--24st + 25 + 16s + 9t - 24st = 25 + 25s2 + 25t2

Therefore two given vectors have the same length.