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Problem 27 Easy Difficulty

Find a unit vector that is orthogonal to both $ i + j $ and $ i + k $.


$$(\mathbf{i}-\mathbf{j}-\mathbf{k}) / \sqrt{3}[\text { or }(-\mathbf{i}+\mathbf{j}+\mathbf{k}) / \sqrt{3}]$$

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Video Transcript

come to this lesson in this lesson, we'll find the in vector. That is also go now, too. Both I have Class J. Then I plus K. They also unit vectors. So here we find that across product mhm of I plus J then I plus key. Okay. Okay. Mhm. Oh, and that is there? Yeah. The first rule as I j and key. Then the 2nd and 3rd as I the component of either is one component of J that is one component of K for the first, then also for this haven't component one zero and one. Okay, so after we're done, we find the magnitude of we divide by the magnitude of that that we get there. Unit Also go now back there of the two. Okay, so here we can find their determinant using the co factor method for the first room to hear this. And that is crossed The Yep. 1001 minus G. So, for J for the Jeff component we delete there. We delete this column and the first rule so that we have 110 and one. I'm for the K. We delete the third column in the first rule so that we have 111 And so Okay, so this becomes this times that this times that so that is one minus zero. So now we have this times that that is one this time start to minus zero. Then that's when we have this time starts. So that is zero minus this times that that is one. Okay, so the whole thing is I minus g, then minus key. Okay, so the victor is I minus j minus K. So we divided by its magnitude. So that is one squared last minus one squared, then plus minus one squared. So that is I minus J minus cape all over the square root of three. Okay, so the unit vector that is also go now to the two unit breakfast. Uh, that is this all over three or the negative of this. Okay, so that is negative. I plus T Lasky, all of us. Current of three. So both vectors how to go now to I plaster, Then I last key. So thanks for your time. This is the end of the lesson.