Let u=j+k and v=i+jsqrt2.
a) Calculate the length of the projection of v in the u direction.
b) Calculate the cosine of the angle between u and v.
Your help is very much appreciated!
u = (1,1) , v = (1,√2)
The projection of v on u
= v.u/|u
= (1 + √2)/√(1^1+1^1) = (1+√2)/√2
or (√2+2)/2
b)
u.v = |u||v|cosØ
1+√2 = √2 √3 cosØ
cosØ = (1+√2)/√6 or (√6 + 2√3)/6
i'm confused on how you got u=(1,1) and v=(1,sqrt2). Shouldn't they each have 3 coordinates?
u=<0,1,1> and v=<1,sqrt2,0> ???
To calculate the length of the projection of v in the u direction, we need to find the scalar projection of v onto u. The scalar projection of v onto u can be found by taking the dot product of v and the unit vector in the direction of u.
a) Let's first find the unit vector in the direction of u:
The direction of u is given by the vector j + k. To find the unit vector in this direction, we divide this vector by its magnitude.
Magnitude of u = sqrt(j^2 + k^2) = sqrt(1^2 + 1^2) = sqrt(2)
Unit vector in the direction of u = (j + k) / sqrt(2) = (1/sqrt(2))j + (1/sqrt(2))k
Now, let's find the dot product of v and the unit vector in the direction of u:
v · (unit vector in direction of u) = (i + jsqrt(2)) · ((1/sqrt(2))j + (1/sqrt(2))k)
= i(1/sqrt(2)) + (jsqrt(2))(1/sqrt(2))
Since j and k are orthogonal vectors, their dot product is zero:
= i(1/sqrt(2))
The length of the projection of v in the u direction is the magnitude of the scalar projection. In this case, it is |i(1/sqrt(2))| = |i|/sqrt(2) = 1/sqrt(2).
b) To calculate the cosine of the angle between u and v, we can use the dot product and magnitudes of u and v:
cos(angle) = (u · v) / (|u| * |v|)
Dot product of u and v: u · v = (j + k) · (i + jsqrt(2))
= ji + j^2sqrt(2) + ki + kjsqrt(2)
= 0 + 1*sqrt(2) + 0 + 1*sqrt(2) = 2sqrt(2)
Magnitudes of u and v: |u| = sqrt(2) and |v| = sqrt(1^2 + (sqrt(2))^2) = sqrt(3)
cos(angle) = (2sqrt(2)) / (sqrt(2) * sqrt(3))
= 2 / sqrt(3)
= (2 / sqrt(3)) * (sqrt(3) / sqrt(3))
= 2sqrt(3) / 3
Therefore, the cosine of the angle between u and v is 2sqrt(3) / 3.