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Homework: Unit 3 HWKScore: 0 of pt27 of 37 (16 complete)Instructor-created questionA force F (3,-2) is applied to a spacecraft with velocity vector(-2,1,0) . Expres...

Question

Homework: Unit 3 HWKScore: 0 of pt27 of 37 (16 complete)Instructor-created questionA force F (3,-2) is applied to a spacecraft with velocity vector(-2,1,0) . Express F as a sum of a vector parallel to and vector orthogonal to vF,uv .O-2,1,0) FzLv

Homework: Unit 3 HWK Score: 0 of pt 27 of 37 (16 complete) Instructor-created question A force F (3,- 2) is applied to a spacecraft with velocity vector (-2,1,0) . Express F as a sum of a vector parallel to and vector orthogonal to v F,uv .O-2,1,0) FzLv



Answers

A force $\mathbf{F}=2 \mathbf{i}+\mathbf{j}-3 \mathbf{k}$ is applied to a spacecraft with velocity
vector $\mathbf{v}=3 \mathbf{i}-\mathbf{j} .$ Express $\mathbf{F}$ as a sum of a vector parallel to $\mathbf{v}$
and a vector orthogonal to $\mathbf{v} .$

We hear told a spacecraft is traveling along Vector v here off. I wrote this in the component for him instead of the I J K form Elector V is three negative 10 and we're applying a force to that vector. And that force is the vector to one negative three. Well, there is some of that force directed along Vector V, and that would be our projection onto vector V off F. And there's some of that force which is at a 90 degree angle to the vector. So these were the two components vector f broken down in reference to Vector V. So they wanted us to write the sum of Vector F has the some of these two vectors one parallel to the which is the projection and one that's perpendicular to the orthogonal to it, which is this blue vector that I have written here, and that blue vector is just gonna be vector f minus the answer for our projection that we have. So we need to find the projection first and we have the formula for that that the projection onto vector V of vector f IHS the dot product of the two divided by the length of vector V squared. And then we take that value times vector B to make it the correct size going in that direction. So f dot v the dot product, multiply your components together and Adam two times three is six plus one times negative one is negative one and negative three times zero is zero. And that's gonna be over the length of the vector V, which is the square root off its components squared at it together. So that would be three squared plus negative one squared plus zero squared, which I'm not gonna bother. And then we square that and we take that times vector V So we have projection is gonna be five over three squared, plus negative. One squared is 10 square with a 10 squared is 10 times vector V, so that's gonna be 1/2 times vector V. So that will be three house negative, 1/2 and zero. So there's our projection, which would have done that in green, so it would have matched the colors. So there is that one. And now this other perpendicular component here f minus the projection that we just calculated. So we're gonna have Tu minus the 3/2 which is 1/2 and then one minus the negative 1/2 which will give us three halves and then negative three minus zero. Which gives us thank you three. So that means that in relation to vector V back to your s would be our projection. Three halves negative 1/2 0 plus the perpendicular component 1/2 three halves and negative three.

So we have about two factors. There s envy and then in order to calculate the dot product, what we need to do is multiply the first row of each vector with the second vote. So eight times 15 and then we add that to the product off. The second round will be tractor. So to sum of minus one, it's that will give us eight times 15 for the first room in blue. And then we add the second row. So too times a negative one. And that gives is 120 for eight times 15 minus two gives us 118 that is foot pounds, so 118.

Hi. So here we have the back to be defined by three I plus four J. And the doctor F. Given by minus six I minus eight J. So we need to find the component of F that is parallel to be. That means that we need to calculate the projection of the vector F along the B. This is because if you think of this problem geometrically here you have your vector B. Here you have your vector F. So the components of F. Of this vector that is horrible to be will be this vector here and you can see that it corresponds to an orthogonal projection. That means that the component of F that is proud to be is the projection of the bacteria F. Uber B. So this as a formula and is equal to the product of F would be divided the product of the wood itself. Time to actually be. Did that product of F would be in this case is equal to -50 And the product of be with itself is the cost to 25. So after replacing the data, you will see that we obtain here -6, I -AJ. What is the meaning of this? This if you can observe is equal to or vector F. That means that F is parallel to the vector B. So we don't have, this is near here, but we have this the better F is parallel to be. And actually, if we consider the court needs, it's something like this. Here is our victory be and or vector F is in the opposite direction but instead is parallel. That means that this picture F doesn't have uh perpendicular components. So this component is zero, I plus zero J. Finally we need to calculate the work done by this force F through the distance beat. But this that product we have already calculated and is -50.

Hi. So for this exercise, let's consider this vector b. and the force vector f given by -0.4 and I bless your .3J. So let's calculate the component of F. That is parallel to be. So this means that we need to calculate the projection of the factor F over B. And we know that this relation this projection is given by the product of F would be divided the dot product of being with itself times the vector B. Let's see what is the value of the product of F would be. And we can see that this is minus one plus one. That is equal to zero. So. Sorry? Yeah. Here have a mistake. Sorry, this And so here is minus 1.2 Plus one. Yeah. And there's the cost to zero. And remember that when the dot product of two vectors is equals to zero is equivalent to say that both vectors are perpendicular. So in this case this projection is zero, so the projection is zero, I bless zero J. There is no component that is parallel to the battery B. Instead, the due to this factor F is perpendicular to be, the perpendicular component will be the factor itself. That means minus your .4, I Plus 0.3. Okay. And the final thing that we need to do is calculate the work done by this spectrum F through the distance B. But we have observed that this the product Is the cost to zero, so the work done is zero.


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