Video Transcript
The diagram shows two vectors, π
and π. Each of the grid squares in the
diagram has a side length of one. Calculate π cross π.
Right, so this is a question about
vector products. We have a diagram showing two
vectors labeled π and π. We are told that the grid squares
have sides of length one. And then weβre asked to calculate
the vector product π cross π of these two vectors.
Letβs start by writing out the
vectors in component form. To do this, we need to find the π₯-
and π¦-components of each vector from the diagram. Letβs add an π₯- and a π¦-axis to
the diagram to make this process a little clearer. We see that vector π extends
positive four units in the π₯-direction and negative four units in the
π¦-direction. If we recall that π’ is the unit
vector in the π₯-direction and π£ is the unit vector in the π¦-direction, then we
can write that the vector π is equal to its π₯-component four multiplied by π’ plus
its π¦-component negative four multiplied by π£. Or more simply, we could write this
as four π’ minus four π£.
For vector π, we see that it
extends negative five units in the π₯-direction and negative one unit in the
π¦-direction. So we can write that π equals its
π₯-component negative five multiplied by π’ plus its π¦-component negative one
multiplied by π£, which we could also write negative five π’ minus π£.
So now we have expressions for both
π and π in component form. The question is asking us to
calculate the vector product π cross π. So letβs recall the definition of
the vector product of two vectors. Letβs define two general vectors
that lie in the π₯π¦-plane and label these lowercase π and lowercase π. Well, weβve used the lowercase
letters to distinguish this general case from our two specific vectors from the
question.
We can write these general vectors
in component form, labeling the π₯-components with a subscript π₯ and the
π¦-components with a subscript π¦. Then, the vector product π cross
π is defined as the π₯-component of π multiplied by the π¦-component of π minus
the π¦-component of π multiplied by the π₯-component of π all multiplied by π€,
which is the unit vector in the π§-direction.
We can use this definition to
calculate the vector product of our two vectors from the question as capital π and
capital π. Itβs going to be important here to
keep track of all the negative signs in the components as we go. We are asked to calculate π cross
π. So the first term is the
π₯-component of π, which is four, multiplied by the π¦-component of π, which is
negative one. Then, from this, we subtract the
second term. This second term is the
π¦-component of π, which is negative four, multiplied by the π₯-component of π,
which is negative five. Then, this whole thing is
multiplied by the unit vector π€.
This first term, four multiplied by
negative one, gives us negative four. The second term, negative four
multiplied by negative five, gives us positive 20. But remember that we are
subtracting this second term from the first. So we have negative four minus 20
all multiplied by π€. Subtracting 20 from negative four,
we get the answer to the question that the vector product π cross π is equal to
negative 24π€.