# Three vectors are v= (3,-1,2), b= (4,2,-5) and n= (1,3,-7). Please prove that they form a closed triangle. What type of triangle is it? There are three vectors. They are v= (3,-1,2), b= (4,2,-5)...

Three vectors are v= (3,-1,2), b= (4,2,-5) and n= (1,3,-7). Please prove that they form a closed triangle. What type of triangle is it?

There are three vectors. They are v= (3,-1,2), b= (4,2,-5) and n= (1,3,-7). Please prove that they form a closed triangle. What type of triangle is it?

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### 2 Answers

If they form a closed triangle, some combination of them must be equal to zero. By observing the given three vectors you can notice that,

`v+n-b = 0`

LHS:

`v+n-b = (3,-1,2) + (1,3,-7) - (4,2,-5) `

`= (4,2,-5) - (4,2,-5)`

`= (0,0,0)`

Therefore the vectors v,n and b form a closed triangle. Actually the order of the vectors are v, n, -b.

To check for the type of the triangle, we can calculate the dot product of the vectors.

v and n,

`v.n = |v|xx|n| xx cos(180-A)`

`-14 = sqrt(14) xx sqrt(59) xx cos(180-A)`

`cos(180-A) = (-14)/(sqrt(14) sqrt(59))`

`180-A = 119`

`A = 61`

v and -b,

`v.(-b) = |v| xx |-b| xx cos(180-B)`

`0 = sqrt(14) xx sqrt(45) xx cos(180-B)`

`cos(180-B) = 0`

`180-B = 90`

`B = 90`

**Therefore the triangle formed is a right angled triangle.**

Since you asked for clarification.

`0 = sqrt(14) xx sqrt(45) xx cos(180-B)`

`cos(180-B) = 0/(sqrt(14) xx sqrt(45))`

`cos(180-B) = 0`

`180-B = cos^(-1) 0`

`180-B = 90`

`B = 90`