We're going to find two unit vectors
orthogonal to both vector v and vector w.
So to do this, we'll first find the cross product
of vector v and vector w.
This'll give us a vector that's orthogonal
to both vector v and w.
And then we'll normalize the vector to determine
one unit vector that's orthogonal
to both vector v and vector w.
And then to find the second unit vector,
we'll determine the unit vector in the opposite direction.
Let's first look at this graphically.
Vector v and vector w are graphed here in red and blue.
We want to determine the two yellow vectors graphed here.
Notice how the unit vectors are opposite vectors,
but they're also perpendicular
to both the red vector and the blue vector,
or vector v and vector w.
So our first step will to be to determine the cross product
of vector v and w.
And we'll set this up as a three by three determinant.
So vector v
cross with vector w
is equal to the three by three determinant
where the first row will be the unit vector i,
the unit vector j, and the unit vector k.
The second row will be the components of vector v.
So we have four, negative one, two.
The third row will be the components of vector w
which are zero, five, and one.
Now to evaluate this determinant,
we use the expansion by minors method given here.
Where this is equal to a two by two determinant
times the unit vector i
minus the two by two determinant times vector j
plus a two by two determinant times vector k.
And now determine the elements
in each two by two determinant.
For this determinant the elements will be determined
by eliminating the row and column of vector i.
So I'll eliminate row one and column one.
So the elements are negative one, two, five, one.
For this two by two determinant we'll eliminate
the row and column of vector j.
So we'll eliminate row one, but now column two.
So we have four, two, zero, one.
And now for the elements here,
we'll eliminate the row and column of vector k.
So we eliminate row one and column three.
So we have four, negative one, zero, five.
Now we'll evaluate each two by two determinant.
And for a quick review, if this is our determinant,
this'll be equal to
a times d minus b times c.
So here we have one times negative one minus two times five.
Well one times negative one is negative one
and minus two times five is 10.
And then we have minus four times one minus two times zero.
Four times one is four minus two times zero is zero.
Times vector j plus here we have four times five
minus negative one times zero,
which would be 20 minus zero times vector k.
So our cross product which we'll call vector a
would be equal to negative 11i
minus four j
plus 20k.
Or in component form we could say
we have an x component of negative 11,
a y component of negative four, and a z component of 20.
And now because vector a is our cross product,
we know this vector is orthogonal
to both vector v and vector w.
And now we'll normalize this to find our first unit vector
that's orthogonal to both vectors.
Let's do this on the next slide.
So for our next step we'll determine
the magnitude of vector a,
which is equal to the square root of negative 11 squared
plus negative four squared plus 20 squared,
which is equal to the square root of 537,
which doesn't simplify.
537 is equal to three times 179.
179 is prime.
So the first unit vector we'll call u sub one
would be equal to
vector a divided by the magnitude of vector a,
which we can also write as one over the square root of 537
times vector a in component form would be
negative 11 comma negative four comma 20.
We could perform square root in multiplication,
but I'll go ahead and leave it in this form.
And our second unit vector will be the unit vector
in the opposite direction.
So vector u sub two would be equal to
the opposite of vector a divided by
the magnitude of vector a
which we can write as negative one
over the square root of 537 times vector a.
Again, we could perform the square root of multiplication,
but we'll leave it in this form.
These would be the two unit vectors
that would be orthogonal to both vector v and vector w.
Which again, going back to our graph,
would be the two unit vectors graphed here in yellow.
Both vectors have a magnitude of one
and are orthogonal or perpendicular to both
the red vector and the blue vector.
I hope you found this helpful.