If a is a vector parallel to the xy-plane and b is a vector parallel to k, determine |a �~ b| when |a| = 3 and |b| = 2.
1. |a �~ b| = −6
2. |a �~ b| = −3�ã2
3. |a �~ b| = 3
4. |a �~ b| = 3�ã2
5. |a �~ b| = 0
6. |a �~ b| = −3
7. |a �~ b| = 6
To determine |a �~ b|, we need to calculate the magnitude of the cross product between vectors a and b.
First, we need to understand the properties of the given vectors. Vector a is parallel to the xy-plane, which means it lies entirely in the xy-plane and has a z-component of 0. Vector b is parallel to the k-axis, which means it has x and y-components of 0.
Since the cross product of two vectors is given by:
a �~ b = (a2b3 - a3b2)i + (a3b1 - a1b3)j + (a1b2 - a2b1)k
In this case, since both a and b have zero components along the z-axis, their cross product will also have a zero component along the z-axis. Therefore, we only need to calculate the cross product of their x and y-components.
Let's calculate the cross product:
a �~ b = (0)(0) - (0)(0)i + (0)(0) - (0)(0)j + (a1b2 - a2b1)k
a �~ b = (a1b2 - a2b1)k
In this particular case, since a lies in the xy-plane, its x and y-components are non-zero, whereas b has zero x and y-components. Therefore, a1b2 = 3(0) = 0, and a2b1 = 0.
Since a1b2 - a2b1 = 0, the cross product a �~ b becomes zero along the k-axis.
Therefore, the magnitude of the cross product |a �~ b| is 0.
Hence, the correct answer is option 5. |a �~ b| = 0.