Final answer:
The magnitude of vectors A and B are, respectively, 7 and 9. The magnitude of A - B is sqrt(222). The dot product of A and B is -46. The cross product of A and B is [-60, 14, 19]. The direction cosine of A with respect to the x-axis is arccos(2/7), and the angle between A and B is arccos((-46) / (7 * 9)).
Step-by-step explanation:
To find the magnitude of a vector, we use the formula:
|A| = sqrt(Ax^2 + Ay^2 + Az^2)
For vector A = [2, -3, 6], the magnitude is:
|A| = sqrt(2^2 + (-3)^2 + 6^2) = sqrt(4 + 9 + 36) = sqrt(49) = 7
Similarly, for vector B = [1, 8, -4], the magnitude is:
|B| = sqrt(1^2 + 8^2 + (-4)^2) = sqrt(1 + 64 + 16) = sqrt(81) = 9
To find the difference between vectors A and B, we subtract their corresponding components:
A - B = [2 - 1, (-3) - 8, 6 - (-4)] = [1, -11, 10]
The magnitude of A - B is:
|A - B| = sqrt(1^2 + (-11)^2 + 10^2) = sqrt(1 + 121 + 100) = sqrt(222)
The dot product of two vectors A and B is given by:
A ⋅ B = Ax * Bx + Ay * By + Az * Bz
For vectors A and B, the dot product is:
A ⋅ B = 2 * 1 + (-3) * 8 + 6 * (-4) = 2 - 24 - 24 = -46
The cross product of two vectors A and B is given by:
A × B = [Ay * Bz - Az * By, Az * Bx - Ax * Bz, Ax * By - Ay * Bx]
For vectors A and B, the cross product is:
A × B = [-3 * (-4) - 6 * 8, 6 * 1 - 2 * (-4), 2 * 8 - (-3) * 1] = [-12 - 48, 6 + 8, 16 + 3] = [-60, 14, 19]
The direction cosine of vector A can be found using:
cosθ = Ax / |A|
θ = arccos(Ax / |A|)
For vector A = [2, -3, 6], the direction cosine with respect to the x-axis is:
θ = arccos(2 / 7)
To find the angle between vectors A and B, we use:
cosθ = (A ⋅ B) / (|A| * |B|)
θ = arccos((A ⋅ B) / (|A| * |B|))
For vectors A = [2, -3, 6] and B = [1, 8, -4], the angle between them is:
θ = arccos((-46) / (7 * 9))