asked 187k views
4 votes
Find the angle θ between a = 3i − k, b = 2i + 2k​

asked
User Pmarcoen
by
8.2k points

1 Answer

3 votes
We can use the dot product formula to find the angle between the vectors:

a · b = ||a|| ||b|| cos(θ)

where ||a|| and ||b|| are the magnitudes of the vectors a and b, respectively.

First, we need to find the magnitudes of the vectors:

||a|| = sqrt(3^2 + 0^2 + (-1)^2) = sqrt(10)
||b|| = sqrt(2^2 + 0^2 + 2^2) = 2sqrt(2)

Then, we can calculate the dot product:

a · b = (3i - k) · (2i + 2k)
a · b = 6 + 0 - 2
a · b = 4

Now we can solve for the angle θ:

a · b = ||a|| ||b|| cos(θ)
4 = sqrt(10) * 2sqrt(2) cos(θ)
cos(θ) = 4 / (sqrt(10) * 2sqrt(2))
cos(θ) = 2 / sqrt(5)

Finally, we can calculate the angle θ:

θ = cos^-1(2 / sqrt(5))
θ ≈ 54.7 degrees

Therefore, the angle between a and b is approximately 54.7 degrees.
answered
User FelixHo
by
8.6k points
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