4. Two vectors are given by Ä = -3+ 4j and B = 2i+ 3j. Find:
a)Ä.B
b) ÄxB
c) the angle between A and B.
a) To find the dot product of two vectors, we multiply their corresponding components and sum them up.
In this case,
Ä.B = (-3)(2) + (4)(3) = -6 + 12 = 6.
b) To find the cross product of two vectors in two dimensions, we take the determinant of the matrix formed by the two vectors' components and the unit vectors i and j.
In this case,
ÄxB = |i j|
|-3 4|
|2 3|
= (4)(3) - (-3)(2) = 12 + 6 = 18.
c) The angle between two vectors A and B can be found using the formula:
cosθ = (Ä.B)/(|Ä|x|B|), where |Ä| and |B| represent the magnitudes of the vectors.
In this case,
|Ä| = sqrt((-3)^2 + (4)^2) = sqrt(9 + 16) = sqrt(25) = 5, and
|B| = sqrt((2)^2 + (3)^2) = sqrt(4 + 9) = sqrt(13).
cosθ = (6)/(5x√13) ≈ 0.511
To find the angle θ, we can take the inverse cosine of cosθ:
θ = cos^(-1)(0.511) ≈ 59.47 degrees.