WEBVTT
00:00:02.730 --> 00:00:09.720
Given that vector π΄ equals one, nine and vector π΅ equals negative four, one, find π΄ minus π΅.
00:00:10.730 --> 00:00:14.860
So in order to actually solve this, what weβre actually going to do is deal with our π₯- and π¦-components separately.
00:00:15.270 --> 00:00:22.830
So when we actually subtract π΅ from π΄ β so our vector π΅ from our vector π΄ β weβre going to deal with our π₯-coordinates and then weβre going to move on to deal with our π¦-coordinates.
00:00:23.540 --> 00:00:35.430
So with that in mind, weβre gonna have π΄ minus π΅ β so vector π΄ minus vector π΅ β is equal to one because thatβs the π₯-component of our vector π΄ minus negative four and thatβs because thatβs the π₯-component of our vector π΅.
00:00:36.410 --> 00:00:38.160
So now we move on to our π¦-components.
00:00:38.930 --> 00:00:45.000
And we have nine minus one because one is actually the π¦-component of our vector π΅.
00:00:45.480 --> 00:00:47.600
Okay, great, so our next stage is to actually simplify.
00:00:48.500 --> 00:00:54.680
So then, our next line in working is that vector π΄ minus vector π΅ is equal to one plus four as our π₯-component.
00:00:55.350 --> 00:01:00.950
And we got one plus four because we had minus a negative and if you minus a negative, this turns positive.
00:01:01.860 --> 00:01:06.050
And then, we moved on to our π¦-component and this stays the same at this point, which is just nine minus one.
00:01:07.270 --> 00:01:18.430
So therefore, we can say that given that our vector π΄ is one, nine and that vector π΅ is equal to negative four, one, vector π΄ minus vector π΅ is gonna be equal to five, eight.
00:01:18.530 --> 00:01:25.490
And we got that because we had one plus four as our π₯-component that gives us five and nine minus one as our π¦-component which gives us eight.