Find the vector...?

Question

Find the vector xi+yj+zk which has a magnitude ot 6cm and makes an angle of 35deg and 80deg with the positive x and y axes respectively.
I don't know how to do it. Mostly cause I don't know which angles it means.

Answer 1

The two angles you're giving is like 2 other vectors of a triangle, you're supposed to get the 3rd one. The magnitude is just 1 vector's length....
Btw vector is the ray/arrow thing, not angle related, just an angle's feets.

Answer 2

The angles are in the plane formed by the vector and the referenced axis. You can find the x-component of the vector from 6*cos(35º), the y-component is 6*cos(80º); from these get the z-component from 6 = √[x^2 + y^2 + z^2]:

z^2 = 36 - x^2 - y^2

z = √[36 - x^2 - y^2]

Answer 3

♦ according to definition dot product of 2 vectors a and b is scalar number
(a·b)=|a|*|b|*cost, where t is angle between them vectors. Thence for a vector r we have got its components as
x=(r·i) = |r|*cos(t1),
y=(r·j) = |r|*cos(t2),
z=(r·k) = |r|*cos(t3);
these cosines are mentioned as directing cosines of a vector r BTW;
♠ according to definition magnitude of vector r is
|r| =√(r·r) =√(x^2 +y^2 +z^2); or;
|r|^2 =(r·r) =x^2 +y^2 +z^2; or;
|r|^2 =(|r|*cos(t1))^2 +(|r|*cos(t2))^2 +z^2, hence
z=±|r|√(1-(cos t1)^2 +(cos t2)^2) two possible values;
now plug in t1=35°, t2=80°, |r|=6cm, and find
x, y, z yourself;
♥ optional: check remarkable property:
(cos t1)^2 +(cos t2)^2 +(cos t3)^2 =1;

Answer 4

the x component is 6 cos 35º and the y component 6cos80º