Video Transcript
Complete the following. If 𝐀 is the vector with components
one, two and 𝐁 equals negative 𝐀, then 𝐀 plus 𝐁 is equal to what.
There are a couple of ways to
answer this question. We could use scalar multiplication
on our vector 𝐀 equals one, two, multiplying it by negative one to get the vector
𝐁. And now recalling that multiplying
any vector 𝐮 by a scalar 𝑘 means we multiply each component of 𝐮 by 𝑘, in our
case, we’re multiplying vector 𝐀 by the scalar negative one. So here 𝑘 is negative one, 𝐮 sub
one is one, and 𝐮 sub two is two. This gives the components of vector
𝐁 as negative one times one, which is negative one, and negative one times two,
which is negative two. And so our vector 𝐁 has components
negative one and negative two.
So now we can find the sum of
vectors 𝐀 and 𝐁 using vector addition, where for vectors 𝐮 and 𝐯 in two
dimensions with components 𝐮 sub one and 𝐮 sub two and 𝐯 sub one and 𝐯 sub two,
respectively, their sum 𝐮 plus 𝐯 has components 𝐮 sub one plus 𝐯 sub one and 𝐮
sub two plus 𝐯 sub two. That is, to sum the vectors, we sum
each pair of components.
So now applying this to our vectors
𝐀 and 𝐁, we have one plus negative one for the first component of the sum and two
plus negative two for the second component, which is actually the zero vector with
components zero, zero. So the sum of vectors 𝐀 and 𝐁,
where 𝐁 is negative 𝐀, is the zero vector.
Now, as an alternative method to
answer this question, we could’ve used just one property of vectors. That’s the additive inverse
property, which tells us that for any vector 𝐮, the sum of 𝐮 with negative 𝐮 is
the zero vector. In our case, we have the sum of
vectors 𝐀 and 𝐁, which since 𝐁 is equal to negative 𝐀 is simply 𝐀 plus negative
𝐀. And by the additive inverse
property, this is equal to the zero vector, which as we know from before is the
vector with components zero, zero.
Hence, if 𝐀 is the vector with
components one and two and 𝐁 is equal to negative 𝐀, then the sum 𝐀 plus 𝐁 is
equal to the zero vector with components zero, zero.