# Question Video: Finding the Magnitude of the Moment of a Force Vector Acting at a Point in Three Dimensions Mathematics

If a force 𝐅 = 6𝐢 − 7𝐣 − 8𝐤 is acting at a point 𝐴(5, −8, 11), find the magnitude of the component of the moment of 𝐅 about the 𝑦-axis.

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### Video Transcript

If a force 𝐅 equal to six 𝐢 minus seven 𝐣 minus eight 𝐤 is acting at a point 𝐴 five, negative eight, 11, find the magnitude of the component of the moment of 𝐅 about the 𝑦-axis.

We can start by drawing a diagram of the situation. We’re told that point 𝐴 has an 𝑥-coordinate of five, a 𝑦-coordinate of negative eight, and a 𝑧-coordinate of 11. That puts it about here. We’re told that a force vector 𝐅 equal to six 𝐢 minus seven 𝐣 minus eight 𝐤 acts at this point. So we can draw this force vector as an arrow on our diagram like this. Now, we’re being asked to find the magnitude of the component of the moment of 𝐅 about the 𝑦-axis. Let’s break this down step by step so we can see exactly what the question is asking here.

Firstly, we can recall that a moment, that is, a rotational force, can be represented by a vector. Generally, we can call this vector 𝐌. Now, we can calculate this vector 𝐌 by finding the vector cross product of the position vector at which the force acts relative to the point we’re calculating moments about and the force vector itself. So calculating this cross product tells us the moment that’s produced by the force 𝐅. In other words, it tells us the moment of 𝐅.

Now, when we calculate the cross product of two three-dimensional vectors, like 𝐑 and 𝐅, the result is a vector with the same number of dimensions. That means that when we calculate the moment vector 𝐌, we obtain a three-dimensional vector. And we can write the components as 𝐌 𝑥 times 𝐢 hat plus 𝐌 𝑦 times 𝐣 hat plus 𝐌 𝑧 times 𝐤 hat. Each of these terms is one of the components of the moment.

We can see that the question asks us for the magnitude of one of the components. Specifically, it’s after the magnitude of the component about the 𝑦-axis. In order to understand what this means, let’s recall that the direction of a moment vector is parallel to the axis of rotation of the moment. This means that the 𝑥-component of a moment vector describes the component of that moment that rotates about an axis pointing in the 𝑥-direction. The 𝑦-component of the moment vector describes the component of that moment that rotates about an axis pointing in the 𝑦-direction. And the 𝑧-component of that moment vector describes the component of the moment that rotates about an axis pointing in the 𝑧-direction.

What this means is that if we calculate a moment about the origin, then, for example, the 𝑥-component of the moment vector describes the component of the moment which causes rotation about the 𝑥-axis itself. Similarly, the 𝑦-component of the moment vector describes the component of the moment which causes rotation about the 𝑦-axis, which is what the question is asking us about.

Finally, the fact that the question just asks for the magnitude of this component just means that we need to give the size of this vector component without writing it as a vector. This bit is fairly straightforward. If the 𝑦-component is 𝐌 𝑦 times 𝐣 hat, then the unit vector 𝐣 hat just tells us the direction of this vector. It points in the 𝑦-direction. And the magnitude of vector is just given by 𝐌 𝑦. So in order to find what this question is looking for, we need to calculate the moment produced by the force 𝐅 about the origin. And in order to do this, we need to find the cross product of these two vectors.

Remember that the vector 𝐑 is the position vector of the point at which the force acts, in this case 𝐴, relative to the point that we’re calculating moments about. In this case, that’s the origin. So the vector 𝐑 is the position vector going from the origin to the point 𝐴. That means that 𝐑, written as the sum of its components, is effectively given by the coordinates of the point 𝐴 multiplied by the respective unit vectors. So we can say 𝐑 is equal to five times 𝐢 hat minus eight times 𝐣 hat plus 11𝐤 hat.

To calculate the cross product of 𝐑 and 𝐅, we need to find the determinant of this three-by-three matrix, where the elements in the top row of the matrix are the three basis vectors that we’re using. The elements in the middle row of the matrix are the three components of the first vector that we’re cross multiplying. In this case, that’s 𝐑. And the elements in the bottom row are the components of the second vector that we’re cross multiplying, in this case 𝐅. Notice that the order of 𝐑 and 𝐅 is important here. The cross product of 𝐑 and 𝐅 is not the same as the cross product of 𝐅 and 𝐑.

There are effectively three parts to calculating the determinant of this matrix. Firstly, we have the unit vector 𝐢 hat multiplied by 𝐑 𝑦 times 𝐅 𝑧 minus 𝐑 𝑧 times 𝐅 𝑦. Next, we subtract the unit vector 𝐣 hat multiplied by 𝐑 𝑥 times 𝐅 𝑧 minus 𝐑 𝑧 times 𝐅 𝑥. And finally, we have the unit vector 𝐤 hat multiplied by 𝐑 𝑥 times 𝐅 𝑦 minus 𝐑 𝑦 times 𝐅 𝑥. These three terms are the 𝑥-, 𝑦-, and 𝑧-components of the moment vector 𝐌. However, because we’re only interested in the magnitude of the component of this moment vector that goes about the 𝑦-axis, that means we’re only interested in this middle term. The magnitude of this component is given by negative 𝐑 𝑥 times 𝐅 𝑧 minus 𝐑 𝑧 times 𝐅 𝑥.

𝐑 𝑥 is the magnitude of the 𝑥-component of the displacement vector 𝐑. In this case, that’s five. And 𝐅 𝑧 is the magnitude of the 𝑧-component of the force vector 𝐅. That’s negative eight. 𝐑 𝑧 is the magnitude of the 𝑧-component of 𝐑, which is 11. And 𝐅 𝑥 is the magnitude of the 𝑥-component of 𝐅, which is six. Altogether, that gives us negative five times negative eight minus 11 times six. Five times negative eight is negative 40, and 11 times six is 66. Negative 40 minus 66 is negative 106. And since there’s a negative sign before the parentheses, this gives us a final answer of 106. And since the question doesn’t specify any particular units of force or displacement, we can just say that our final answer is 106 units of moment.