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 = i + 2j, w = 3i + 6j

Accepted Answer

Hello, today we're going to be determining the projection of vector A onto vector B. Then we're going to find 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 go ahead and start with the first part of the problem. We're going to be calculating the projection of A on to be. Now the definition of a projection of A on to B is given to us as the dot product of the two vectors A and B over the magnitude of B squared multiplied by the vector of B. So there's a couple of things that we'll need to calculate in order to get the projection of A onto B. First, let's calculate the dot product of vector A and vector B. Now vector A is given to us as 27 I plus two J and vector B is given to us as 81 I plus six J. In order to take the dot product, we're going to take the product of the individual components and add those products together. So we're going to take the product of the I components of A and B which is going to be 27 multiplied by 81. Then we're going to add the product of the J components, which is going to be two multiplied by six multiplied by 81 will give us the value of 2187 and two multiplied by six will give us the value of 12. Adding these two values together will give us a final value of 2199. So this is going to be the dot product of A and B. Next, we're going to need to calculate the magnitude squared of vector B. Now in order to calculate the magnitude squared of any vector, that is just going to be the square of the individual components or the sum of the square of the individual components. So the magnitude of B squared is going to be defined as 81 squared plus six squared, 81 squared will give us the value of 6561 and six squared will give us the value of 36. Adding these two values together will give us a final total of 6597. So that's going to be the magnitude of B squared. If we were to plug this in to the projection, we can rewrite the projection of A on to be as 2199 over 6597 multiplied by the B vector, which is given to us as 81 I plus six J. Now the fraction 2199 over 6597 is going to simplify to give us one third. So the projection of A on to B can be rewritten as one third multiplied by 81 I plus six J. In order to simplify the projection, we're going to distribute the scalar one third into the vector 81 I plus six J. And doing so is going to give us a projection of 27 I plus two J. This is going to be the vector that represents the projection of A onto B. Now let's move on to the second part of this question. We are asked to find two vectors A sub one and a sub two where a sub one is parallel to vector B and as a two is perpendicular to vector B, let's start by finding the vector A sub one. Now, if we want a vector A sub one to be parallel to vector B in two dimensions, the definition of a sub one is just going to be the projection of A onto B. So that means a sub one is going to equal to the projection of A onto B which is going to equal to 27 I plus two J. That is going to be the value of a sub one. Now, we want to find a sub two where a sub two is perpendicular to vector B. In order to get this vector, we'll take our A vector and subtract the vector A sub one from that vector. So a sub two is going equal to the A vector which was given to us as 27 I plus two J. And we're subtracting the A sub one vector which is 27 I plus two J. We're first going to distribute negative one into the A sub one vector. And that is going to leave us with 27 I plus two J minus 27 I minus two J. If we were to add the I components and J components together, what we are left with is zero I plus zero J. And this is going to be the value of the A sub two vector. So just to summarize the projection of A on to B was calculated as 27 I plus two J. In order to get a vector A sub one to be parallel to vector B, that vector is just going to equal to the projection which is 27 I plus two J. And in order to get a vector A sub two to be perpendicular to the vector B, we just took the difference of the vector A and the A sub one vector which gave us a zero vector. And with that being said, the answer to this problem is going to be a. 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 + 2j, w = 3i + 6j

Key Concepts

Vector Projection

Dot Product

Vector Decomposition

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