Question
Lineari ze the Ollown_ equhun, So that' V.' is # dpendkat Nariable and + is "Ousr indferdint Variabl . Vc Voe-t/ec
Lineari ze the Ollown_ equhun, So that' V.' is # dpendkat Nariable and + is "Ousr indferdint Variabl . Vc Voe-t/ec
Answers
Find and $a v$ when $u=-1, v=2$ if $w=x y+\ln z$ $x=v^{2} / u, y=u+v, z=\cos u$
Hello, guys. So, for the following, uh, problem. We got these two vectors. So you is to hear mhm 21 01 minus one on the other, vector V is minus 23102 So the are two vectors on. We got the following equation. A U plus the he is equals to minus 88 Three miners. One seven. Okay, so what happened? Well, from this, uh, equality here, we got some linear equations. So the equations are too. Yeah, to a minus. To B is equals to minus eight. Okay, then we got a plus three b is equals to eight. B is equal to three on a is equal to minus one on minus A plus to be is equals. Thio to 27 Torrey here, this is equals to seven. Okay, so we got this equation. Let's say one. This is 2345 Andi. Well, you cannot do that in three on before we got our dissolution for this system or some indication, toe off. What are dissolution of this system of equations? So we is going to be three on a is equal to minus one. So what remains for this exercise is to check that. Evidently, this, uh, these values for a M B hold for the for each of the equations. So there are many equations is 12 and 12 and five. So let's check. So let's check what happened. So in the one we got to a minus to B is equals toe minus eight. So let's replace the values here. We got two minus one minus two three. This part here is six on this is minus two. So we obtain minus a. So they're equal. So it holds for the first equation. Then let consider the second equation, which is a plus three beat plus three b is equals to eight. So again, here a is minus one plus three times we on b is three. So this is minus one last nine on this is just eight, So this quantities are equal, so it also holds for the second equation. The last one is the five. The question number five. So this is minus A plus to be is equals to seven. Okay, again, from here we go that minus minus one plus to be that is three. But it's is just one plus six and this s seven. So these two quantities here are equal, so it also holds for the fifth equation. So that's it. The result is that a is equal to minus one on B is equal to three.
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Okay. He is a vector so that the dot product with V with ia zero V with J zero and be with K zero, find V. So let's make our vector V A comma Beacom asi. So we can write out that dot products as we already know what I is. It's 100 We know J 010 and then K is 001 So we can do the dot products with any of these. So if our dot product V dot A has to equal zero, they would have to equal zero, V. Dot J equals B. But if it has to equal zero, B equals zero. And again be dot K would equal C. And so C has to equal zero. So the only vector in which this works is R zero vector.