find the angle between these lines
[x,y,z] = [3,4,1] + s[-1,5,-2]
[x,y,z] = [8,5,-9] + t[2,3,-6]
recall that you can find the angle θ between vectors u and v using
u•v = |u|*|v|*cosθ
so, just plug and chug:
-2+15+12 = √30√49*cosθ
cosθ = 25/(7√30) = .652
θ = 0.86 = 49.3°
thank you !!
To find the angle between two lines, we can use the dot product formula:
cos(theta) = (u · v) / (|u| * |v|)
where u and v are the direction vectors of the lines. In this case, we need to find the direction vectors of both lines.
The first line can be represented as:
x = 3 - s
y = 4 + 5s
z = 1 - 2s
To find the direction vector, we can ignore the constants and write the coefficients of s as the vector components. Therefore, the direction vector of the first line is [-1, 5, -2].
Similarly, the second line can be represented as:
x = 8 + 2t
y = 5 + 3t
z = -9 - 6t
The direction vector of the second line is [2, 3, -6].
Now we can find the angle between these lines by substituting the direction vectors into the dot product formula:
cos(theta) = (u · v) / (|u| * |v|)
cos(theta) = ([-1, 5, -2] · [2, 3, -6]) / (sqrt((-1)^2 + 5^2 + (-2)^2) * sqrt(2^2 + 3^2 + (-6)^2))
To calculate the dot product:
([-1, 5, -2] · [2, 3, -6]) = (-1 * 2) + (5 * 3) + (-2 * -6) = -2 + 15 + 12 = 25
To calculate the magnitudes:
|[-1, 5, -2]| = sqrt((-1)^2 + 5^2 + (-2)^2) = sqrt(1 + 25 + 4) = sqrt(30)
|[2, 3, -6]| = sqrt(2^2 + 3^2 + (-6)^2) = sqrt(4 + 9 + 36) = sqrt(49) = 7
Substituting the values back into the formula:
cos(theta) = 25 / (sqrt(30) * 7)
To find the angle theta, we take the inverse cosine (arccos) of cos(theta):
theta = arccos(25 / (sqrt(30) * 7))
Using a calculator, we can compute the value of theta to get the angle between the lines.