In Exercises 42–43, find projᵥᵥv. Then decompose v into two vecto...

Question

In Exercises 42–43, find projᵥᵥv. Then decompose v into two vectors, v₁ and v₂ where v₁ is parallel to w and v₂ is orthogonal to w.

v = -2i + 5j, w = 5i + 4j

Accepted Answer

Hello. Today we're going to be using the two given vectors to solve the following problem. So we are told that vector A is equal to 12 I plus five J. And we are told that vector B is equal to two I plus six J. The first part of the problem asks us to calculate the projection of vector A onto vector B. Now the projection of vector A onto vector B is written in this format where A is in the parentheses of the projection and B is the subscript of the projection. Now, the projection of A on to B is defined as the dot product of A and B divided by the magnitude of B squared multiplied by the vector B. So in order to calculate the projection, we're going to need to calculate the dot product of A and B and we're going to need to calculate the magnitude of B squared. Let's start with the dot product A and B. Now, in order to get the dot product of A and B, we'll need to multiply the I components together, multiply the J components together, then we'll need to take the sum of both of the products. So the magnitude of A on to of A and B is equal to 12, multiplied by two plus five, multiplied by six, 12, multiplied by two will give us the value of 24 and five multiplied by six will give us the value of 30 and 24 plus 30 will give us a final sum of 54. So we have the dot product of A and B. Now we'll need to calculate the magnitude of B squared. Now, the magnitude of B squared is the sum of the components of B squared. So the magnitude of B squared will be defined as two squared plus six squared, two squared is equal to four and six squared is equal to 36. Finally, four plus 36 will give us a final sum of 40. So now that we have the dot product of A and B and we have the magnitude of B squared, we can use these values to calculate the projection of A on to B. So the projection of A onto B is equal to 54/40 multiplied by vector B which is given to us as two I plus six J. The value 54/40 will simplify to give us 27/20. And that is being multiplied by the vector B. Now, in order to simplify this projection, we'll need to distribute 27/20 into the components of vector B and that will give us a final projection of 27/10, I plus 81/10 J. This will be the vector that represents the projection of A onto B. Now that we've calculated the projection, we can move on to the second part of the problem. Now, the second part of the problem asks us to calculate two vectors A one and A two vector A one will be parallel to vector B and vector A two will be perpendicular to vector B. Let's start by calculating vector A one. Now in order for a one to be perpendicular to vector B vector A one will equal to the projection of A on two B. But we've already calculated the projection of A on two B. It is 27/10, I plus 81/10 J. What this means is that vector A one is equal to 27/10 I plus 81/ J. This will be the vector that is parallel to vector B. Next, we'll need to calculate vector A two and A two must be perpendicular to vector B. In order to get this vector, we'll take the difference of vector A and vector A one. Now recall that vector A is given to us as 12 I plus five G. So if we wanted to take this difference, we're going to write vector A which is 12 I plus five J and subtract vector A one which is 27/10, I plus 81/10 J. In order to simplify this vector, we'll distribute negative one into vector A one that will leave us with 12 I plus five J minus 27/10, I minus 81/10 J. Finally, if we were to subtract the I components together 12 I minus 27/10, I will give us a final value of 93/10 I. And if we were to subtract the J components together five J minus 81/10 J will give us a final value of negative 31/10 J. That means vector A two is equal to 93/10, I minus 31/10 J. So just to summarize vectors A and B were used to calculate projection A on to B which is 27/10 I plus 81/10 J. The vector A one which is parallel to vector B is equal to the projection of A on to B and the vector A two which is perpendicular to vector B is the difference of vector A minus A one which gives us the final value of 93/10, I minus 31/10 J. 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.

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In Exercises 42–43, find projᵥᵥv. Then decompose v into two vectors, v₁ and v₂ where v₁ is parallel to w and v₂ is orthogonal to w. v = -2i + 5j, w = 5i + 4j