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

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 = i + 3j, w = -2i + 5j

Accepted Answer

Hello, today we're going to solve the following information for the given vectors. So the first part of the question asks us to calculate the projection of vector A onto vector B. Now the projection of vector A on vector B is written in this notation and in order to get the projection of A on to B that is going to equal to the dot product of A and B divided by the magnitude of B squared multiplied by the vector B. So if we want to calculate the projection of vector A on vector B, we have two things we first need to calculate. First, let's start by calculating the dot product of A and B. Now vector A is given to us as 11 I minus J and vector B is given to us as three I plus nine J. In order to calculate the dot product between A and B, we'll need to multiply the I components together and we need to multiply the J components together. And the dot product is going to be the sum of both of those products. So the dot product of A and B is going to be multiplied by three plus negative one multiplied by nine 11, multiplied by three will give us the value of 33 and negative one multiplied by nine will give us the value of negative nine 33 minus nine will give us a final sum of 24. So that means the dot product of A and B is equal to 24. Next, let's calculate the magnitude of B squared. In order to calculate the magnitude of B squared, we'll take the sum of the components of, of B squared. So the magnitude of B squared is equal to three squared plus nine squared. Three squared gives us the value of nine and nine squared. Gives us the value of 81 and nine plus will give us a final sum of 90. So that means the magnitude of B squared is equal to 90. Now that we have the dot product of A and B and the magnitude of B squared. If we were to plug this into the projection, the projection of A onto B is equal to 24/90 multiplied by the vector B which is three I plus nine J 24/90 will simplify to give us 4/15. So we have 4/15, multiplied by three I plus nine J. In order to simplify the projection, we're going to distribute 4/15 into the vector B and that is going to give us a final vector of 4/5 I plus 12/5 J. This is going to be the vector that represents the projection of A onto B. Now that we have the projection of A on to B, we can move on to the next part of the question. The next part of the question asks us to find two vectors A one and A two A one is going to be parallel to vector B and A two will be perpendicular to vector B. Let's start by finding a vector A one per parallel to vector B. Now in order to get a parallel vector A one will equal to the projection of A on two B and we've already calculated the projection of A on to B. It's 4/5 I plus 12/5 J. That means the vector A one parallel to vector B is equal to 4/5 I plus 12/ J. This will be the answer for a one. Next, we need to calculate a vector A two and A two must be perpendicular to vector B. In order to get a perpendicular vector A two, we're going to take the difference of vector A and vector A one. Now recall that vector A is given to us as 11 I minus J and vector A one is 4/5 I plus 12/5 J. That means vector A two will equal to 11 I minus J minus 4/5 I plus 12/5 J. In order to simplify A two, we're first going to distribute negative one into the vector A one that is going to leave us with 11 I minus J minus 4/5, I minus 12/5 J. The next thing we're gonna want to do is we're going to want to reorder the terms, let's put the I terms together and the J terms together that will leave us with 11 I minus 4/5, I minus J minus 12/ J. Now 11 I minus 4/5, I will give us a final value of 51/5 I and negative J minus 12/5 J will give us a final value of negative 17/5 J. This is going to be the vector A two which will be perpendicular to vector B. So just to summarize, we've calculated the projection of A on to B which is 4/5 I plus 12/5 J. The vector A one parallel to vector B is the same value as the projection of A on to B and A two which is a vector perpendicular to B is equal to 51/5, I minus 17/5 J. With that being said, the answer to this problem is going to be C. 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.

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 = i + 3j, w = -2i + 5j

Key Concepts

Vector Projection

Dot Product

Vector Decomposition

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