In Exercises 33–38, find projᵥᵥ v. Then decompose v into two vect...

Question

In Exercises 33–38, find projᵥᵥ v. Then decompose v into two vectors, v₁ and v₂, where v₁ is parallel to w and v₂ is orthogonal to w.v = 3i - 2j, w = i - j

Accepted Answer

Hello, today we're gonna be determining the projection of vector A on A vector B. Then we're going to calculate two vectors A sub one and a sub two where A sub one is parallel to vector B and A sub two is perpendicular to vector B. Let's start by calculating the projection of A onto B. The projection of A on to B is defined as the dot product of A and B over the magnitude of B squared multiplied by the vector of B. So there are a few things that we need in order to calculate the projection of A on to B. Let's start by calculating the dot product of A and B vector A is given to us as six, I minus seven J and vector B is given to us as I minus four J. In order to calculate the dot product, we're going to take the product of the I N J components and we're going to add the two products. So the dot product of A and B is going to be defined as six multiplied by one plus negative seven multiplied by negative four, six multiplied by one will give us the value of six and negative seven multiplied by negative four will give us the value of positive 28 and six plus 28 will give us the final value of 34. Next, let's calculate the magnitude of B squared. The magnitude of B squared is just going to be the sum of the squares of the components of B or in other words, the magnitude of B squared is defined as one squared plus negative four squared. One squared is going to evaluate to give us one and negative four squared will evaluate to give us positive 16 and one plus 16 will give us the value of 17. So if we were to plug in the dot product and the magnitude of B squared back into the projection, the projection of A onto B is defined as 34/17 multiplied by the B vector which was given to us as I minus four J 34/17 will simplify to give us two. So we are left with two multiplied by I minus four J. We're going to distribute two into the vector B and that is going to give us a final projection of A onto B as two I minus eight J. Now that we have the projection of A on to B, we're going to calculate the two vectors A sub one and a sub two. Remember a sub one must be parallel to vector B and A sub two. Must be perpendicular to vector B. In order to get a sub one, A sub one parallel to vector B is just going to equal to the projection of A onto B. Now, we calculated this projection earlier which was two I minus eight J. So a sub one is equal to the projection of A on to B which is equal to two I minus eight J. Now we're going to calculate the second vector A sub two and A sub two must be perpendicular to vector B. In order to get this vector, we're going to take the given a vector and subtract the vector A sub one from the vector A. So vector A was given to us as six I minus seven J and the vector A sub one is equal to the projection of A on to B which is two I minus eight J. In order to get a sub two, we're going to distribute negative one into two I minus eight J. That will leave us with six I minus seven J minus two, I plus eight J. Then we're going to add the respective I and G components together. And that is going to give us an ace up two value of four I plus G. So just to summarize the projection of A on two, B is two I minus eight J A parallel vector to B is just a projection of A on to B which is two I minus eight J and A perpendicular vector to B is the difference of A minus a sub one which is four I plus J. And with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 33–38, find projᵥᵥ v. Then decompose v into two vectors, v₁ and v₂, where v₁ is parallel to w and v₂ is orthogonal to w.v = 3i - 2j, w = i - j

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