Find the vector, not with determinants, but by using properties of cross products.
$ (i + j) \times (i - j) $
welcome to this lesson. In this lesson, you'll find a cross border of I J and I minus J. Norwood. They're determined. Mattered byword properties. So, yeah, if you have a better days, E one eight to a three. Yeah, from the Director B, that is B one B two, B three. The cost for that. Their cost product is E three. Uh, the cross board that is eight to be three minus B two 83 than E three. Be one minus B three minus B three a one. Then the final part is a one B two minus B 182 Okay, so here. Well, let a B the air would be one 10 because there's no component that the cave part for A And also for this one, the B would be one negative one and zero. Okay. So that a Cosby would be. So you have eight to that is one times B three, that is zero the minus. B two, that is negative. One, an e three. That is sterile. Okay, so that is one part. The first back, the f part. Then they gave bad. We have a three that is zero then be one that is one minus B three. That is zero than a three. That is one. Okay, so that is also gone. Then we have a one. That is one be too. That is negative one, then minus. We have a one that be one that is one then eight to that is also one. Okay. And I represent the cave path. So here the air Crosby would be zero for the i component. Therefore the chief component than negative one minus one for the cave components. So here we have a Crosby, that is Yeah. Zero. I last zero j minus two key. Okay, so it Cosby would be negative. Two k. All right, so this is a time short time the end of the lesson.