### Video Transcript

If π and π are any two vectors,
is the following algebraic expression true? π minus π equals negative π
minus π.

There are two ways we can go about
verifying whether this expression is true graphically or algebraically. Letβs start by doing this
graphically.

In order for this statement to be
generally true, it must be true for any two vectors π and π that we could possibly
choose. So, it doesnβt really matter what
π and π are because it should be true no matter what. Letβs choose to define π and π
like this. First, weβll work out the resultant
vector of the left-hand side of this expression, π minus π. To do this, we need to subtract
vector π from vector π. We might recall that we can easily
add vectors by lining them up tip to tail. But how do we subtract vectors?

Well, we can think about this
expression another way. Just like with normal numbers,
subtracting a vector is the same as adding its negative. So, π minus π can be rewritten as
π plus negative π. Negative π is the vector that has
the same magnitude as π but that points in the opposite direction. So, we would draw negative π like
this. Now all we need to do is add
negative π to π by positioning them tip to tail like so. The result is equal to this vector
here, which weβll call π sub L to remind us itβs from the left-hand side of the
equation.

Next, letβs look at the right-hand
side of the equation, negative π minus π. Letβs start by finding the vector
inside the brackets, π minus π. Just like before, we can rewrite
this as π plus negative π, and we can draw the vector negative π like so. Now we add this to vector π, which
leaves us with this resultant vector. However, we have to be careful not
to forget the minus sign in front of the brackets. This tells us we need to reverse
the direction of this vector like so. This gets us our final vector π
sub R.

Now, letβs compare our two vectors,
π L and π R. We can see that theyβre the
same. In other words, we see that the
resultant vector from the left-hand side of the equation is equal to the resultant
vector from the right-hand side of the equation. Hence, we have verified that this
expression is true for these two vectors that we chose. However, we could have approached
this question in a different way that wouldnβt involve us drawing out so many
vectors. If we approach this question
algebraically, we can test whether this expression is true for any two vectors
beyond just two random vectors that we think up.

Recall that a vector can be
expressed algebraically in terms of its horizontal and vertical components, π₯ and
π¦, and the horizontal and vertical unit vectors, π’ hat and π£ hat. We can rewrite vector π as π
equals π΄ sub π₯ π’ hat plus π΄ sub π¦ π£ hat and vector π as π΅ sub π₯ π’ hat plus
π΅ sub π¦ π£ hat. Letβs sub these expressions into
the left-hand side of the equation we want to verify. We find that π minus π is equal
to π΄ sub π₯ π’ hat plus π΄ sub π¦ π£ hat minus π΅ sub π₯ π’ hat plus π΅ sub π¦ π£
hat, which equals π΄ sub π₯ π’ hat plus π΄ sub π¦ π£ hat minus π΅ sub π₯ π’ hat
minus π΅ sub π¦ π£ hat. Now, letβs gather like terms and
factor out the coefficients of π’ hat and π£ hat. This leaves us with π΄ sub π₯ minus
π΅ sub π₯ π’ hat plus π΄ sub π¦ minus π΅ sub π¦ π£ hat.

Next, letβs think about the
right-hand side of the expression, negative π minus π. If we substitute in our expressions
for π and π, we find this is equal to the negative of π΅ sub π₯ π’ hat plus π΅ sub
π¦ π£ hat minus π΄ sub π₯ π’ hat plus π΄ sub π¦ π£ hat. If we expand out all the brackets,
paying attention to the two minus signs, we find this simplifies to negative π΅ sub
π₯ π’ hat minus π΅ sub π¦ π£ hat plus π΄ sub π₯ π’ hat plus π΄ sub π¦ π£ hat. Again, if we gather like terms and
factor out the coefficients of π’ hat and π£ hat, weβre left with π΄ sub π₯ minus π΅
sub π₯ π’ hat plus π΄ sub π¦ minus π΅ sub π¦ π£ hat as our expression for the
right-hand side. This is exactly the same as we
found for the left-hand side of the equation. So, we know the equation must be
correct.

We have now verified the expression
π minus π equals negative π minus π using two different methods. The answer to this question is
therefore yes. This expression is true for any two
vectors.