Please find the coordinates of this vector d=(a,b,c), that makes an angle of 30 degrees with the y-axis, 60 degrees with he z axis and is perpendicular to the x-axis. Is the answer distinct?   

Expert Answers
mlehuzzah eNotes educator| Certified Educator

First, picture, for a moment, in 2D, the following four vectors:

<1,1>, <-1,1> , <-1,-1>, <1,-1>

They each make a 45 degree angle with the x-axis (and a 45 degree angle with the y-axis). Moreover, we could stretch our vector:
<2,2> , <3,3>, etc, also make a 45 degree angle with each of the x and y axes.

So if we were looking for a vector (in 2D) that made a 45 degree angle with the x and y axes, the answer would not be unique.

The nonuniqueness comes from two places:

1) To be 45 degrees from the x-axis means to make a 45 degree angle with EITHER of <1,0> or <-1,0>. That is, you have two choices for the vector you make an angle with, corresponding to the positive and negative parts of the axis

2) the vector can be made a different length, without changing the angle

Ok, now to your actual question:

The x-axis is the collection of the multiples of <1,0,0>

To be perpendicular to this, we need `<a,b,c>*<1,0,0> =0`

So we must have `a=0`

To figure out the other coordinates, we need to use the formula:

`A*B=|A||B| "cos" theta`

We want <0,b,c> to make a 30 degree angle with EITHER <0,1,0> or <0,-1,0>

Using our formula:

`<0,b,c>*<0,+-1,0> =(sqrt(b^2+c^2))(1)( "cos" 30)`

`+-b=(sqrt(b^2+c^2))( (sqrt(3))/(2))`

Now, because we may scale the vector to have any length, we can stretch (or shrink) the vector so that b has length 1:





Thus we get the following four vectors:





As well as all of their multiples

And, in fact, if we stretch the vector by a factor of `(sqrt(3))/2`, we get the more familiar: