### Video Transcript

The forces π
one is equal to negative 10π’ minus seven π£, π
two is ππ’ minus π£, and π
three is five π’ plus π minus 10π£ act on a particle, where π’ and π£ are two perpendicular unit vectors. Given that the forces resultant π is negative 13π’ minus three π£, determine the values of π and π.

In this example, we want to determine the unknown values of π and π, appearing in the components of the forces π
two and π
three, respectively. These two forces, together with π
one, act on a particle with a given resultant force. Now recalling that the resultant force acting on any body is the sum of all the forces, then the resultant force of the three forces acting on this particle is π is equal to π
one plus π
two plus π
three. That is negative 10π’ minus seven π£ plus ππ’ minus π£ plus five π’ plus π minus 10π£.

And now collecting like terms, that is, the coefficients of π’ and π£, we have negative 10 plus π plus five multiplied by π’ and negative seven minus one and π minus 10 multiplied by π£. And this simplifies to π minus five times π’ plus π minus 18 times π£. But weβre already told that the resultant force is negative 13π’ minus three π£. And equating coefficients, this must mean that π minus five is negative 13 and π minus 18 is negative three.

Now, if π minus five is negative 13, then π is negative 13 plus five; thatβs adding five to both sides. And so, π is negative eight. And similarly, if π minus 18 is negative three, then adding 18 to both sides gives us π is negative three plus 18. And thatβs equal to 15. Hence, the values of π and π are π is negative three and π is 15.