𝐴𝐵𝐶𝐷 is a rectangle, where 𝐜
hat is a unit vector perpendicular to its plane. Find the cross product of vector
𝐂𝐌 and vector 𝐂𝐁.
We recall that the cross product of
two vectors 𝐚 and 𝐛 is equal to the magnitude of vector 𝐚 multiplied by the
magnitude of vector 𝐛 multiplied by sin of angle 𝜃 multiplied by the unit vector
𝐧. 𝜃 is the angle between the two
vectors, and the unit vector 𝐧 is perpendicular to vectors 𝐚 and 𝐛. When dealing with a two-dimensional
geometric shape, this will be perpendicular to the plane.
In this question, we need to
calculate the cross product of vector 𝐂𝐌 and vector 𝐂𝐁. As the magnitude of any vector is
its length, then the magnitude of 𝐂𝐁 is equal to 44, as the sides 𝐃𝐀 and 𝐂𝐁 on
the rectangle are equal to 44 centimeters. We can see from the diagram that
the magnitude of vector 𝐂𝐌 will be equal to one-half of the magnitude of vector
As triangle 𝐴𝐵𝐶 is a right
triangle, we can use the Pythagorean theorem to calculate the length of 𝐂𝐀. The Pythagorean theorem states that
𝑎 squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the length of the
longest side or hypotenuse. The magnitude of vector 𝐂𝐁
squared plus the magnitude of vector 𝐁𝐀 squared will be equal to the magnitude of
vector 𝐂𝐀 squared. Substituting in our values from the
diagram, we need to calculate 44 squared plus 33 squared. This is equal to 3025. Square rooting both sides of this
equation gives us that the magnitude of vector 𝐂𝐀 is equal to 55. The length of the diagonal of the
rectangle from point 𝐶 to point 𝐴 is 55 centimeters. One-half of 55 is 27.5. Therefore, the magnitude of vector
𝐂𝐌 is 27.5.
We now need to calculate the angle
between our two vectors, which we will call 𝛼. We can do this using our trig
ratios. And we know that sin of angle 𝛼 is
equal to the opposite over the hypotenuse. The length of 𝐁𝐀 is equal to 33
centimeters, and the length of 𝐂𝐀 is equal to 55 centimeters. Therefore, sin 𝛼 is equal to 33
over 55. By dividing the numerator and
denominator by 11, this simplifies to three-fifths.
When dealing with a cross product,
we measure the angle in a counter- or anticlockwise direction. If we wanted to calculate 𝐂𝐁
cross 𝐂𝐌, then the angle 𝜃 would be as shown in the diagram such that sin 𝜃 is
equal to three-fifths. We don’t want to calculate this,
however. We want to calculate 𝐂𝐌 cross
𝐂𝐁. This means that the angle will be
negative 𝜃. We know that the sine function is
odd, which means that sin of negative 𝜃 is equal to negative sin 𝜃. This means that the value of sin 𝜃
in our example will be negative three-fifths.
This leads us to an interesting
rule about the vector or cross product. Vector multiplication is not
commutative. 𝐚 cross 𝐛 is not equal to 𝐛
cross 𝐚. However, as the sine function is
odd, 𝐚 cross 𝐛 is equal to negative 𝐛 cross 𝐚. This means that in our question,
𝐂𝐌 cross 𝐂𝐁 is equal to negative 𝐂𝐁 cross 𝐂𝐌.
Using the formula for the cross
product, 𝐂𝐌 cross 𝐂𝐁 is equal to 44 multiplied by 27.5 multiplied by negative
three-fifths multiplied by the unit vector 𝐜. This is equal to negative