### Video Transcript

The force π
equals three π’ minus
four π£ is acting on the point π΄ zero, two. Given that π΅ equals two, three and
πΆ equals negative two, one, determine the moment of this force about both points π΅
and πΆ. Which of the following would you
conclude about the line of action of the force π
? (A) The line of action of π
passes
through the point πΆ. (B) The line of action of π
bisects the line segment π΅πΆ. (C) The line of action of π
is
parallel to the line π΅πΆ. Or (D) none of the answers are
correct.

Letβs begin with a rough diagram of
the scenario. We have the three points π΄ zero,
two; π΅ two, three; and πΆ negative two, one. And we have the force π
equals
three π’ minus four π£ acting on the point π΄. Recall that the moment π of a
force π
acting from a point π about a pivot point π is given by π« cross π
,
where π« is the vector π to π.

In this case, we are asked to
determine the moment of π
about the points π΅ and πΆ. We therefore need to find the
vectors π« π΅, which is equal to π΅ to π΄, and π« πΆ, which is equal to the vector
πΆ to π΄. So π« π΅ is equal to the position
vector of π΄ zero, two minus the position vector of π΅ two, three, which comes to
negative two, negative one. Similarly, π« πΆ is equal to the
position vector of π΄ minus the position vector of πΆ, which is zero, two minus
negative two, one, which comes to two, one.

Notice that this is exactly the
same vector but with the opposite sign to π« π΅. So, when we take the cross product,
the moment will be the same but with the opposite sign. This allows us to immediately
conclude that the line of action of π
bisects the line π΅πΆ. Since the moment is the same
magnitude about both points, the perpendicular distance of the line of action of the
force must be the same for both points. And since the sign of the moment is
opposite for the two points, the line of action must pass between the two points
rather than on the same side if it were parallel to the line between the two
points.

Now we need to determine the moment
of the force about the points. The moment π π΅ about the point π΅
is equal to π« π΅ cross π
, which is given by the determinant of the three-by-three
matrix π’, π£, π€, negative two, negative one, zero, three, negative four, zero. Both of these vectors are in the
π₯π¦-plane and have a π€-component of zero. Therefore, only the π€-component of
their cross product will be nonzero.

Taking the determinant of this
matrix by expanding along the top row gives us 11π€. Since π« πΆ is equal to negative π«
π΅, the moment about the point πΆ will just be equal to the negative of the moment
about the point π΅. So this gives us our final
answer. The moment about the point π΅ is
equal to 11π€, and the moment about the point πΆ is equal to negative 11π€.