Find the vector cross product of two vectors C=2i-j and D=i+2j-3k. What is the angle between the vectors
To find the vector cross product of two vectors C and D, you can use the following formula:
C x D = (C_y * D_z - C_z * D_y) i + (C_z * D_x - C_x * D_z) j + (C_x * D_y - C_y * D_x) k
where C_x, C_y, and C_z represent the components of vector C, and D_x, D_y, and D_z represent the components of vector D.
Given C = 2i - j and D = i + 2j - 3k, we can substitute these values into the formula to find the vector cross product:
C x D = ((-1 * (-3)) - (0 * 2)) i + ((0 * 1) - (2 * (-3))) j + ((2 * 2) - ((-1) * 1)) k
= (3 - 0) i + (0 + 6) j + (4 + 1) k
= 3i + 6j + 5k
Now, to find the angle between two vectors C and D, you can use the dot product formula:
C · D = |C| * |D| * cos(θ)
where C · D represents the dot product of vector C and D, |C| and |D| represent the magnitudes of the vectors C and D, and θ represents the angle between the two vectors.
Given C = 2i - j and D = i + 2j - 3k, we can first find the magnitudes of these vectors:
|C| = sqrt((2^2) + (-1^2)) = sqrt(4 + 1) = sqrt(5)
|D| = sqrt((1^2) + (2^2) + (-3^2)) = sqrt(1 + 4 + 9) = sqrt(14)
Next, we can find the dot product of C and D:
C · D = (2 * 1) + (-1 * 2) + (0 * -3) = 2 -2 + 0 = 0
Substituting these values into the dot product formula, we can solve for the angle θ:
0 = sqrt(5) * sqrt(14) * cos(θ)
Since the dot product is 0, the angle between the two vectors is 90 degrees or π/2 radians.