### Video Transcript

π
one equals ππ’ plus π£ and π
two equals ππ’ minus five π£, where π
one and π
two are two forces acting at the points π΄ three, one and π΅ negative one, negative one, respectively. The sum of the moments about the point of origin equals zero. The sum of the moments about the point πΆ one, two also equals zero. Determine the values of π and π.

Recall that the moment π of a force π
about a point π is equal to the cross product π« cross π
, where π« is the vector from π to the point that the force acts from, π΄. The point π can be the origin or any other point in space. If it is the origin, then π« is simply the position vector of the point π΄.

In this question, we are asked to determine the values of two unknowns, π and π. To find the value of two unknowns, we will need two separate simultaneous equations. In this case, our first equation will be given by the sum of the moments about the origin equaling zero. And our second equation will be given by the sum of the moments about the point πΆ equaling zero.

Letβs begin with the first equation about the origin. The position vector π of the point π΄ is three, one. And the position π of the point π΅ is negative one, negative one. Therefore, the moment π one of the force π
one is equal to π cross π
one. And the moment π two of the force π
two is equal to π cross π
two. Taking this first cross product, π cross π
one, we have the determinant of the three-by-three matrix π’, π£, π€, three, one, zero, π, one, zero. Both forces and position vectors lie in the π₯π¦-plane. Therefore, everything is two-dimensional and we have a zero column in the π€ column.

Because of this, only the π€-component of both moments will be nonzero. This makes sense because, remember, a cross product is always perpendicular to both vectors. And since both vectors are in the π₯π¦-plane, the result must be parallel to the π§-axis. Taking this determinant then by expanding along the top row leaves us with just three minus π times π€. Similarly, for π two, we have the determinant of the matrix π’, π£, π€, negative one, negative one, zero, π, negative five, zero. Once again, both of these vectors are in the π₯π¦-plane, and the π€ column is zero. So only the π€-component will be nonzero. Taking this determinant by expanding along the top row gives us five plus π times π€. Adding these together gives us eight minus π plus ππ€.

The question tells us that the sum of these two moments is equal to zero or the zero vector. For a vector to be equal to the zero vector, all of its components must also be zero. Therefore, the scalar quantity eight minus π plus π must be equal to zero. This, therefore, gives us our first equation: eight minus π plus π equals zero.

Now, we need to find a second equation from the sum of the moments about the point πΆ one, two being equal to zero. Since we are finding the moment about the point πΆ instead of the origin, the vectors π and π will be different. π will be given by the position vector of the point π΄ three, one minus the position vector of the point πΆ one, two. This comes to two, negative one. Likewise, π will be equal to the position vector of the point π΅ negative one, negative one minus the position vector of πΆ one, two, which comes to negative two, negative three.

Again, the moment π one of the force π
one about the point πΆ will be given by π cross π
one. And the moment π two of the force π
two about the point πΆ will be given by π cross π
two. Taking this first cross product gives us the determinant of the matrix π’, π£, π€, two, negative one, zero, π, one, zero. Again, everything is in the π₯π¦-plane. And we are left with only the π€-component, which gives us two plus π times π€. Likewise, the cross product π cross π
two gives us the determinant of the matrix π’, π£, π€, negative two, negative three, zero, π, negative five, zero. And this evaluates to 10 plus three π times π€.

Again, the question tells us that the sum of these two moments, π one plus π two, is equal to the zero vector. So we have 12 plus π plus three π times π€ equals the zero vector. Again, this means the π€-component must also be equal to zero. And this gives us our second simultaneous equation: 12 plus π plus three π equals zero.

We can now solve these two equations to find the values of π and π. If we add the first equation to the second equation, the π and negative π will cancel, and we are left with 20 plus four π equals zero. Solving for π gives us π equals negative five. Substituting this value of π into the first equation gives us eight minus π minus five equals zero. And solving for π gives us π equals three.