### Video Transcript

What is the torque about the origin
of the force 5.0π’ minus 2.0π£ plus 1.0π€ newtons if it is applied at the point
whose position is π« equals negative 2.0π’ plus 4.0π£ meters?

In this problem, weβre given a
force that acts at a point, and weβre told the force in terms of its components:
5.0π’ minus 2.0π£ plus 1.0π€ newtons. We can picture this force acting in
a three-dimensional space being applied at a point whose distance from the origin
negative 2.0π’ plus 4.0π£ meters.

Because this force has π’, π£, and π€ components, itβs a three-dimensional force. And we expect that our resulting
torque will also be in three dimensions. We want to solve for the torque,
letβs call that π, and recognize that itβs a vector. Having both magnitude and
direction, weβll use the Greek letter lower case π to represent torque. And letβs begin by recalling the
definition of what torque is.

Torque, π, is defined as the cross
product of π«, our position vector and π
, our force vector. Since weβre combining π« and π
through a cross product, we know that the result, torque, will itself be a vector
having magnitude and direction. It will be important for us to
recall the formula that teaches us how to compute a cross product between two
vectors, so letβs recall that relationship now. The cross product between two
three-dimensional vectors, π and π, is equal to these π’, π£, and π€ components
that we see written here. This equation may look long and
confusing, but there is an order and a method to it.

First of all, notice that for every
pair of components of the vectors π and π, say this first pair π΄ sub π¦ times π΅
sub π§, the next pair that we see is the reverse of that with subscripts π΄ sub π§
and π΅ sub π¦. And this pattern goes on for the π£
and π€ components as well. Also notice that for each one of
the three components π’, π£, and π€, the components of the vectors π and π that
are used to calculate that term are perpendicular or orthogonal to those
components.

For example, with the π£ component
of π cross π, normally associated with the π¦-value of those vectors, we see the
only components of π and π involved are π₯ and π§ orthogonal to π¦. This three-dimensional cross
product formula is useful to have memorized, but if not, we can always look it
up. Letβs now apply this cross product
formula to our scenario using π« and π
in place of π and π, respectively. And just a quick note, order is
very important with cross products. For example, π« cross π
is not the
same as π
cross π«. In general, those two cross
products are not equal.

So we want to find the cross
product of π« cross π
which is equal to the torque. To do that, letβs apply our cross
product equation to the π« vector and π
vector given to us in the problem
statement.

As we work along, π«, our vector
for position, is represented by π in the cross product formula. And π
, our vector for force, is
represented by π. Letβs begin with the π’th component of π« cross π
. π« cross π
is equal to π sub π¦
times πΉ sub π§ minus π sub π§ times πΉ sub π¦. For the π£ component of π« cross π
,
we have π sub π§ times πΉ sub π₯ minus π sub π₯ times πΉ sub π§. And for the π€th component, we
have π sub π₯ times πΉ sub π¦ minus π sub π¦ times πΉ sub π₯.

Weβre now ready to begin to plug in
the values for these components of π« and π
, working off of the π« and π
we have
defined in the problem statement. Letβs start with the π’th
component of π« cross π
. π sub π¦ is 4.0. πΉ sub π§ 1.0. π sub π§ is zero, because there is
no π€th component of π«. And πΉ sub π¦ is negative 2.0. 4.0 times 1.0 minus 0 is 4.0.

Now we move on to the π£th component of π« cross π
. π sub π§ is zero, because there is
no π€th component of π«. π sub π₯ is negative 2.0. And πΉ sub π§ is 1.0. Zero minus negative 2.0 times 1.0
all together equals 2.0. Now moving on to the π€th
component, π sub π₯ is negative 2.0, πΉ sub π¦ is negative 2.0, π sub π¦ is 4.0,
and πΉ sub π₯ is 5.0. Negative 2.0 times negative 2.0
minus 4.0 times 5.0 all together equals negative 16.

Since weβve solved for π« cross π
,
by the definition of torque weβve also solved for the torque. We find that the torque in this
scenario is equal to 4.0π’ plus 2.0π£ minus 16π€. Since we combine units of force
that is Newtons and units of distance, that is meters to create torque, the units of
torque are Newton meters.