# Answer to Question #5214 in Linear Algebra for Jill

Question #5214

Use the cross product to find the sine of the angle between the vectors u = (2,8,-2) and v = (2,8,2) .

Expert's answer

u = (2,8,-2) and v = (2,8,2)

Let's use the formula:

|uXv| =

|u|*|v|*sinA.

uXv = (u2v3 - u3v2)*i + (u3v1 - u1v3)*j + (u1v2 - u2v1)*k

=

(8*2 + 2*8)*i + (-2*2 - 2*2)*j + (2*8 - 8*2)*k = 32i - 8j + 0k,

|uXv| =

sqrt(32^2 + 8^2) = sqrt(1088),

|u|*|v| = sqrt(2^2 + 8^2 + 2^2)*sqrt(2^2 + 8^2

+ 2^2) = 72.

So, sinA = sqrt(1088) / 72 = 0.458.

Let's use the formula:

|uXv| =

|u|*|v|*sinA.

uXv = (u2v3 - u3v2)*i + (u3v1 - u1v3)*j + (u1v2 - u2v1)*k

=

(8*2 + 2*8)*i + (-2*2 - 2*2)*j + (2*8 - 8*2)*k = 32i - 8j + 0k,

|uXv| =

sqrt(32^2 + 8^2) = sqrt(1088),

|u|*|v| = sqrt(2^2 + 8^2 + 2^2)*sqrt(2^2 + 8^2

+ 2^2) = 72.

So, sinA = sqrt(1088) / 72 = 0.458.

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