# VectorsVectors (0,2,1) and (-2,4,5) and a third vectors form a triangle. What is the height of this triangle?

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### 1 Answer

The "height" of the triangle depends on which side you declare your base.

Picture any (non equilateral) triangle on a piece of paper. Rotate it so that a different side is the base. You'll observe that the height depends on which side is the base.

That said, let's just figure out the side lengths to start with:

`<0,2,1>` has length `sqrt(0^2+2^2+1^2)=sqrt(5)`

`<-2,4,5>` has length `sqrt((-2)^2+4^2+5^2)=sqrt(45)`

The final side must be given by the vector <-2-0,4-2,5-1> = <-2,2,4>

So it has length

`sqrt((-2)^2+2^2+4^2)=sqrt(24)`

So let's think of this as a triangle ABC with sides a, b, c

side a is across from angle A

side b is across from angle B

side c is across from angle C

side a has length `sqrt(5)`

side b has length `sqrt(45)`

side c has length `sqrt(24)`

Now comes the ambiguous part. Any one of these can be placed as the base. I am going to take b as the base, and find the height, but you can take any of these to be the base, and you will get different heights.

We can calculate angle A (or any of the angles) using the law of cosines:

`a^2=b^2+c^2-2bc "cos" A`

`5=45+24-2sqrt(24*45) "cos" A `

`2sqrt(24*45) "cos" A = 64`

`"cos" A = (32)/(6sqrt(30))`

`"cos" A = .9737`

`A=.2299` ` `(in radians)

Now, to get the height of the triangle, consider the right triangle made out of c, part of b, and the height of the triangle, h. (We "drop a perpendicular" from angle B, at the top of the triangle)

This right triangle has a hypotenuse of `sqrt(24)` and an angle of .2299 radians

So we use:

`"sin" A = ("OPP")/("HYP")`

`"sin" .2299 = (h)/(sqrt(24))`

`h = sqrt(24) "sin" .2299 = 1.116`

So the height of the triangle (assuming the long side is the base) is 1.116