a. The dot product A.B:
A.B = (3i + 2j - k) . (5i + 5j)
A.B = 3(5) + 2(5) + (-1)(5)
A.B = 15 + 10 - 5
A.B = 20
b. The projection of A onto B:
The projection of A onto B is given by the formula: projB(A) = (A.B / ||B||^2) * B
where ||B|| is the magnitude of B.
First, we need to find the magnitude of B:
||B|| = sqrt(5^2 + 5^2) = sqrt(50) = 5sqrt(2)
Then, we can calculate the projection of A onto B:
projB(A) = (A.B / ||B||^2) * B
projB(A) = (20 / (5sqrt(2))^2) * (5i + 5j)
projB(A) = (20 / 50) * (5i + 5j)
projB(A) = (2/5) * (5i + 5j)
projB(A) = 2i + 2j
c. The angle between A and B:
The angle between A and B is given by the formula: cos(theta) = (A.B) / (||A|| ||B||)
First, we need to find the magnitudes of A and B:
||A|| = sqrt(3^2 + 2^2 + (-1)^2) = sqrt(14)
||B|| = sqrt(5^2 + 5^2) = sqrt(50) = 5sqrt(2)
Then, we can calculate the cosine of the angle:
cos(theta) = (A.B) / (||A|| ||B||)
cos(theta) = 20 / (sqrt(14) * 5sqrt(2))
cos(theta) = 4 / (sqrt(14) * sqrt(50))
cos(theta) = 4 / (sqrt(700))
cos(theta) = 0.171
Finally, we can calculate the angle between A and B:
theta = cos^-1(0.171)
theta = 1.383 radians (approximately 79.3 degrees)
d. A vector of magnitude 2 in the XY plane perpendicular to B:
Since B is in the XY plane (it has no component in the Z direction), any vector perpendicular to B must have a non-zero Z component. Also, since we want a vector of magnitude 2, we can assume that its Z component is 0. Therefore, we need to find a vector in the XY plane (i.e., with Z = 0) that is perpendicular to B and has magnitude 2.
One such vector is the cross product of B with the unit vector in the Z direction:
C = B x k = (5i + 5j) x k
C = -5i + 5j
To get a vector of magnitude 2, we can scale this vector by a factor of 2/|C|:
v = (2/|C|)C
v = (2/5) * (-5i + 5j)
v = -2i + 2j
Therefore, a vector of magnitude 2 in the XY plane perpendicular to B is -2i + 2j.