### Video Transcript

The forces π
one equals two π’ plus two π£, π
two equals ππ’ plus nine π£, and π
three equals nine π’ plus ππ£ act on a particle, where π’ and π£ are two perpendicular unit vectors. Given the forcesβ resultant π equal to two π’ minus six π£, determine the values of π and π.

Letβs begin by recalling what we actually mean by a resultant force. The resultant force is the overall force acting on the object. Here, itβs the sum of π
one, π
two, and π
three. Now, weβre actually given that the resultant of these forces is two π’ minus six π£. Well, since the resultant is the sum of the three forces, we can say that π
sub one plus π
sub two plus π
sub three must be equal to π. Weβre going to replace each force as given in the question. And we see that two π’ plus two π£ plus ππ’ plus nine π£ plus nine π’ plus ππ£ must be equal to two π’ minus six π£.

Next, weβre going to collect together the π’-components and, separately, the π£-components of each force on the left-hand side. The π’-components are two, π, and nine. And the π£-components are two, nine, and π. And so, we can say that two plus π plus nine π’ plus two plus nine plus π π£ must be equal to two π’ minus six π£. Now, of course, for the vector on the left-hand side to be equal to the vector on the right-hand side of our equation, we know that the individual components must themselves be equal. Equating the components in the π’-direction, and we get two plus π plus nine equals two.

Similarly in the π£-direction, our equation is two plus nine plus π equals negative six. We can simplify our first equation, and we get π plus 11 equals two. Then, we solve for π by subtracting 11 from both sides. And we find π to be equal to negative nine. Similarly, our other equation becomes 11 plus π equals negative six. To solve this equation for π, weβre going to subtract 11 from both sides. And we get π equals negative 17. So, given the information about our three forces and their resultant, we found π is equal to negative nine and π is equal to negative 17.

And in fact, we could check our answers by substituting π equals negative nine and π equals negative 17 back into our original forces. Doing that and then finding the sum of the three forces, we do indeed find that itβs equal to two π’ minus six π£, which we saw is equal to π.