Video Transcript
Consider the two vectors 𝐑 equals
three 𝐢 plus two 𝐣 and 𝐒 equals five 𝐢 plus eight 𝐣. Calculate 𝐑 cross 𝐒.
Well, we see this exercise is about
computing a vector product between these two vectors 𝐑 and 𝐒. We’ve been given those two vectors
in component form. And we see that each one has an
𝐢-component, that is a component along the 𝑥-axis, as well as a 𝐣-component, some
component along the 𝑦-dimension. We see then that both of these
vectors, 𝐑 and 𝐒, lie in the 𝑥𝑦-plane. In fact, if we were to sketch out
𝑥- and 𝑦-axes, then we can draw in vectors 𝐑 and 𝐒 on this graph.
Vector 𝐑 extends three units in
the positive 𝑥-direction and two units in the positive 𝑦, giving a vector like
this. While vector 𝐒 extends five units
in the positive 𝑥-direction and then eight units in the positive 𝑦-direction. And that vector looks like
this. Now, to calculate this vector
product, 𝐑 across 𝐒, we’ll want to recall the mathematical form of the vector
product of two vectors that lie entirely in the 𝑥𝑦-plane like 𝐑 and 𝐒 do.
If we call two general vectors that
both are constrained to lie in the 𝑥𝑦-plane 𝐀 and 𝐁. Then the vector product of 𝐀 and
𝐁, also called the cross product of 𝐀 and 𝐁, is equal to the 𝑥-component of 𝐀
times the 𝑦-component of 𝐁 minus the 𝑦-component of 𝐀 times the 𝑥-component of
𝐁. Note that a vector product of two
vectors results in a vector. That is, it has both magnitude as
well as direction. And since our vectors 𝐀 and 𝐁
both lie in the 𝑥𝑦-plane, their vector product points perpendicularly to that, in
the 𝐤-direction.
Let’s now use this relationship
along with 𝐑 and 𝐒 written in their component form to calculate 𝐑 cross 𝐒. This vector product is equal to the
𝑥-component of 𝐑 times the 𝑦-component of 𝐒 minus the 𝑦-component of 𝐑 times
the 𝑥-component of 𝐒. And this will also point in the
𝐤-direction, either positive or negative. Looking at these different terms in
our parentheses, we can see that the 𝑥-component of 𝐑 is three, that the
𝑦-component of 𝐒 is eight and then the 𝑦-component of 𝐑 is two and the
𝑥-component of 𝐒 is five.
When we substitute in these values,
our next task is to calculate their result. Three times eight is 24, and two
times five is 10. And then, 24 minus 10 is equal to
14. So, our vector product points 14
units in the positive 𝐤-direction. So, if we drew in a 𝑧-axis on our
graph, where that axis pointed out of the screen at us, then this vector product
would point 14 units toward us along that direction. This is the vector product of 𝐑
and 𝐒.
