### Video Transcript

End π΄ of line segment π΄π΅ is at
negative six, seven and π΄π΅ has midpoint π· negative seven, one. If the line of action of the force
π
equals negative two π’ minus six π£ bisects π΄π΅, determine the moment of the
force π
about point π΅.

Remember, we can calculate the
moment of a force π
taken about some point π by finding the cross product of π«
with vector π
, where π« is the position vector of π΄, which is the point of
application of the force. But in this case weβre not given
the point at which the force is acting. We are, however, told that the line
of action of the force π
bisects π΄π΅. This tells us then that the line of
action of the force passes through the midpoint π· negative seven, one. Now, we know that as long as the
point lies in the same line of action of the force, then the vector moment π is
independent of the initial point. And so we can calculate the moment
by considering that initial point to be at π· negative seven, one.

Since weβre not working with the
origin, weβre going to replace the vector π« with the vector ππ. Weβre finding the moment of π
about π΅, assuming that the force is acting at π·. So letβs begin by finding the
vector ππ. Now, we know that the midpoint of
line segment π΄π΅ is point π·. So if we plot point π΄ and point π·
on a coordinate plane, this then allows us to show that the magnitude, the length,
of the vector ππ must be equal to the length of the vector ππ. But of course these are acting in
opposite directions, so we can actually say that the negative vector ππ is equal
to the vector ππ.

Then, we can find the vector ππ
by subtracting the vector ππ from the vector ππ. So thatβs the vector negative
seven, one minus the vector negative six, seven. Subtracting the individual
components, and we get negative one, negative six. Then, the vector ππ is the
negative of this vector. So we multiply through by negative
one, and we find vector ππ is simply one, six.

And we now have enough to calculate
the moment of our force. Since the force π
is negative two
π’ minus six π£, the moment is the cross product of the vector one, six with the
vector negative two, negative six. And using the 2D definition of the
cross product, we multiply one by negative six and then subtract six times negative
two. And thatβs all multiplied by the
unit vector π€. Well, one times negative six minus
six times negative two is positive six. And so we found the moment of our
force π
about point π΅ to be six π€.