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# Find two unit vectors orthogonal to both $j - k$ and $i + j$.

## $$\frac{\sqrt{3}}{3} \mathbf{i}-\frac{\sqrt{3}}{3} \mathbf{j}-\frac{\sqrt{3}}{3} \mathbf{k} \quad \text { and } \quad-\frac{\sqrt{3}}{3} \mathbf{i}+\frac{\sqrt{3}}{3} \mathbf{j}+\frac{\sqrt{3}}{3} \mathbf{k}$$

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in this problem, we're trying to find two unit vectors that are orthogonal to A and be you guessed it. This is a cross product problem To find two unit vectors that are orthogonal to two other vectors. We can just start by finding any vectors that are orthogonal to those two vectors, which means calculating the cross products between them in the second part will just scale those down until their unit factors. So if we want to find the cross product of A and B, it helps. If you write a zero, I plus one, j minus one K And be as one, I plus one J plus zero K. Then our cross product is going to let's solve this using the same method in our textbook, Although we could use properties of unit factors here. But if we ignore our first column, We have one time 0 Right is negative one times 1 makes negative one times one. Hi minus. Then we'll ignore our second column And we'll look at zero times 0 minus negative one times one. That's going to be zero times 0 minus negative one times one plus. And if we ignore our third column, this is turning into a bit of a mess. Look at zero times 1 -1 times one. Crohn's one -1 times one. Okay, if we simplify all of this, that's going to give us one minus zero is zero minus negative one times one. That's -2. one eye or just one for our first uh component of this sector, zero minus negative one times one is again one except there's a negative sign out front. So this is actually negative one. And lastly zero minus negative one. Sorry, 0 -1 times one as negative one. So this is one vector that is orthogonal to a cross B. That is orthogonal to both. A and B. Excuse me. But how long is it? Well, the length let's call that L. Is just going to be the square root of each term square. So that'll be one squared plus negative one squared plus negative one squared Which is just the square root of three. So what we can do is take every term in this vector and divided by three. So the first vector that we're looking for will be one over route three, negative one over Route three and negative one over root three. That's one unit vector. That's orthogonal to both A and B. How do we find another one? Well, if we have one vector going this way, let's just look at the other vector going the other way. If we make every term in this negative, that gives us negative one over route three positive one over Route three positive one over Route three. And we could simplify this. However we want if we want to use i J k notation, that's fine. If we want to uh complete the square so that we I don't have any radicals on the bottom. That's fine as well. But both of these will give us a mathematically correct answer. Thanks for.

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