Question

Consider the x–axis as representing east, the y–axis as north and z–axis as vertically upwards. Give the vector representing each of the following points.
a) 5 m north east and 2 m up
b) 4 m south east and 3 m up
c) 2 m north west and 4 m up

Answer 1

Final Answer :
$a) \: \: \frac{5}{ \sqrt{2} } i + \frac{5}{2} j + 2k \\ b) \: 2 \sqrt{2} i - 2 \sqrt{2} j + 3k \\ c) - \sqrt{2} i + \sqrt{2} j + 4k$
^_^
I MADE THE GRAPH IN 2-D FOR BETTER UNDERSTANDING OF QUESTION. •_^

Steps:
1) General form of any point in 3-D is given as
R = xi +yj +zk
where x, y, z are corresponding co -ordinates .

a) In 5m North East and 2m up :
Z-component , z = 2 ( above the plane )

North East direction is 45° North of East.
Hence , looking at pic 1 :
$x = 5 \cos(45 \degree) = \frac{5}{ \sqrt{2} } \\ y = 5 \sin(45 \degree) = \frac{5}{ \sqrt{2} }$
Hence,
Position of Point in 3-D : (5/√2,5/√2, 2)

b) In 4m South East, 3m up
z-component = 3
=> z = 3

South East Direction is 45° South of East.
Look at pic 2 :
$x = 4 \cos(45 \degree) = 2 \sqrt{2} \\ y = - 4 \sin(45 \degree) = - 2 \sqrt{2}$
Y is negative since, South is in downward direction.
Therefore,
Position of Point in 3-D = (2√2,-2√2,3)

c) In 2m North west, and 4m up.
z - component, z = 4
North West direction is 45° North of West.
Hence,
$x = - 2 \cos(45 \degree) = - \sqrt{2} \\ y = 2 \sin(45 \degree) = \sqrt{2}$

Hence, Position of Point in 3- D is
(-√2,√2 , 4)