It's calculus, help!!!

Consider the vectors a = <1,1,0> and b = <0,1,-1>. Find a vector c which is perpendicular to a and makes an angle of 60 degree with b.

How do i solve it? THANK YOU

 

I solve it myself, haha
a = <1, 1, 0 >
b = <0, 1, -1>

Let
c = <d, e, f>

The dot product of orthogonal vectors is zero.

a • c = <1, 1, 0 > • <d, e, f> = d + e = 0
e = -d

So we have c = <d, -d, f>.
____________

b • c = <0, 1, -1> • <d, -d, f> = -d - f

Take the magnitude of vectors b and c.

| b | = √[0² + 1² + (-1)²] = √(0 + 1 + 1) = √2
| c | = √[d² + (-d)² + f²] = √(2d² + f²)

The dot product can also be represented another way.

b • c = | b | | c | cos60°

because 60° is the angle between vectors b and c.

b • c = | b | | c | cos60°
-d - f = (√2) [√(2d² + f²)] (1/2)
-d - f = [√(2d² + f²)] (1/√2)

Square both sides.

(-d - f)² = (2d² + f²) (1/2)
2(d² + 2df + f²) = 2d² + f²
2d² + 4df + 2f² = 2d² + f²
4df + f² = 0
f(f + 4d) = 0

f = 0, -4d
The solution f = 0 is extraneous and therefore rejected.

f = -4d

So

c = <d, -d, -4d>

Let d = 1.

Then
c = <1, -1, -4>
________

Test.

a • c = <1, 1, 0> • <1, -1, -4> = 1 - 1 + 0 = 0
So vectors a and c are perpendicular.

Now vectors b and c.

b • c = <0, 1, -1> • <1, -1, -4> = 0 - 1 + 4 = 3
| b | = √2
| c | = √(2d² + f²) = √(2*1 + 4²) = √(2 + 16) = √18

b • c =? | b | | c | cos60°
3 = (√2)(√18) cos60°
3 = (√36) (1/2) = 6 * (1/2)
3 = 3

So vector c = <1, -1, -4> is a solution.

 

well done, doesnt that feel better than someone you telling you the answer :biggthumpup:

 

That's vectors lol, not calc.

 

its in my calculus book, so ya, haha

 

Vector Calculus, by E. Marsden?

 

it is a part of calculus.... i took 3 semester of calculus and this was in Calc III

 

Cool, I'm doing chapter 7 and 8 right now

 

They incorporated vectors in so people could get a more broader understanding of math. But like I said it is not calculus, part of it no doubt.

 

hahaha
seems i cant remember anything i did during my degree
math is such a stranger now