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Homework Help: Grade 12U Calculus - Algebraic Vectors+Cross&Dot Product

  1. May 22, 2008 #1
    Hey guys, I know i've left it kind of late, but I have a calc. unit test tomorrow at 10AM. I was doing the assigned review, and some extra questions of top of it, and these ones are still giving me trouble.

    Any help at all you could offer would be greatly appreciated, as I have to pull my mark up back to an 80-82% (i'm sitting at a 77% right now, we just finished a tough unit)

    The problem statement, all variables and given/known data
    #3a. Find the dot product of the two vectors u and v where u=(3,-4,1) and v=(2,1,5)
    b. Find the angle between u and v.

    For this question, i copied part B from a friend, in hopes of understanding it, but I'm still not sure how to find the angle.
    #7. The cosine of the angle between a and b is 4/21. Find p, if a=6i+3j-2k and b=-2i+pj-4k

    For this one, i'm not even sure if I'm on the right track. I think I just confused myself
    #9. Calculate the dot product of 4x-y and 2x+3y if |x|=3, |y|=4 and the angle between x and y is 60degrees.

    I'm not sure where to go on this one either.
    #12. A triangle has vertices A(-1,3,4), B(3,-1,1) and C(5,1,1)
    a) show that the triangle is right-angled
    b) Calculate the area of the triangle
    c) Calculate the perimeter of triangle ABS
    d) Determine the forth vertex needed to complete a rectangle

    Whenever I try to graph 3D shapes, I get confused. I have my drawing below, but I still can't visualize how it looks. Also how do I get the area of the triangle without the height?
    #13. Find the projection of u=(17,-3,8)
    a) onto each of the coordinate axes
    b) onto each of the coordinate planes

    I remember doing the projection stuff, its applications of dot product if i'm not mistaken, but we never did an example like this, only ones projecting one given vector on another (both with numbers)
    #14. Use the cross product to find the area of the triangle whose vertices all lie in the xy-plane at coordinates A(-7,3,0), B(3,1,0), and C(2,-6,0)

    No idea what to do here :(
    #1 What can you conclude about the vectors u and v if
    c) u x v = 0
    d) |u x v| = |u||v|
    e) (u x v) dot u = 0
    f) (u x v) x u = 0

    These communication questions always get me, again i'm stumped.
    #2 Given u=6i+3j+2k and v=-3i+4j+k find
    d) a unit vector perpendicular to both u and v

    Where did I go wrong in my calculations here?
    #3a) Draw x-, y- and z-axes and make a sketch of
    i) the position vector OP of the point P(3,-2,5)
    ii) the projection of OP onto the z-axis
    iii) the projection of OP onto the xy-plane
    b) Determine the magnitudes of the projections in a, parts ii and iii.

    This one is similar to #13, but like I said before, we've never done an example of a projection onto an axis or plane.

    The attempt at a solution

    http://img399.imageshack.us/img399/6462/tehinterwebsjb5.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. May 22, 2008 #2
    For 3b, everything is correct until the last line. It should be theta = inversecosine (-3 / sqrt(780)) which is about 96.2 degrees. The solution is based on the definition of the dot product (u dot v = mag u times mag v times cos of the angle between them) and the formula for calculating the dot product when the vectors are given in rectangular component form (u dot v = uxvx + uyvy + uzvz).
  4. May 22, 2008 #3
    For 7, you're close: 3p - 4 = mag a times mag b times (4/21). Calculate mag a and mag b, plug in and solve for p.
  5. May 22, 2008 #4
    For 9, no clue. The notation makes no sense to me.
  6. May 22, 2008 #5
    For 13a, a unit vector parallel to the x-axis is (1,0,0), to the y-axis is (0,1,0), the z- axis is...
  7. May 22, 2008 #6
    For 14, form two vectors using two pairs of the vertex coordinates. The area is (1/2) the magnitude of of their cross product.
  8. May 22, 2008 #7
    Why (4/21)? Where did that come from?
  9. May 22, 2008 #8
    #9. Calculate the dot product of 4x-y and 2x+3y if |x|=3, |y|=4 and the angle between x and y is 60degrees.

    I'm not sure where to go on this one either.

    = 8|x|^2+10x.y-3|y|^2
    and x.y. = |x|.|y|.cos theta ...

    I would be surprised if you don't see this in the test =P
  10. May 22, 2008 #9
    #14. Use the cross product to find the area of the triangle whose vertices all lie in the xy-plane at coordinates A(-7,3,0), B(3,1,0), and C(2,-6,0)

    No idea what to do here :(

    Also, a really good question =P

    Area = 0.5*side1*side2*sin theta
    so get two vectors say AB and AC
    and you know ABxAC = ... ? (magnitude)
  11. May 22, 2008 #10
    3a) You calculated correctly, but do not write <6,-4,-5> in the middle step, which implies that the dot product is a vector. Just write 6 + (-4) + (-5) = -3.

    3b) You've (almost) written down the basic equation useful for dot products. It can be proven that

    [tex]\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta[/tex]​

    where [itex]\theta[/itex] is the angle between the two vectors if you put their tails together. It is important to realize that you do not need the bars around the dot product -- a dot product is not a vector. The rest is just solving for [itex]\theta[/itex]. When you have time, you can find and read the proof for this equation from your textbook or even in the Wikipedia entry for dot product.

    Like edziura says, make sure you can solve it correctly.

    7) You've almost got it. It's a test on dot and cross products, so when you see cosine of an angle between the two vectors, your mind should jump to dot products. Midway through, you got to

    [tex]3p - 4 = |\vec{a}||\vec{b}|\cos(79^\circ)[/tex]​

    Then, in the next step, the cosine disappeared. What should the cosine be? A few lines down, you suddenly decided to set 3p - 4 = 0. Why? Use the equation I pointed out above and plug in values for each of the quantities. Sure, [itex]|\,\vec{b}\,|[/itex] will still have a p in it; so solve for p.

    9) The dot product is distributive. Thus, we have

    [tex](\vec{r} + \vec{s}) \cdot (\vec{u} + \vec{v}) = \vec{r}\cdot\vec{u} + \vec{s}\cdot\vec{u} + \vec{r}\cdot\vec{v} + \vec{s}\cdot\vec{v}[/tex]​

    just as when you are multiplying two binomials with FOIL. You should be able to solve from there using the information given. How do you come up with this line of thought? Well, you are given [itex]|\,\vec{x}\,|[/itex], [itex]|\,\vec{y}\,|[/itex], and the angle between them. This is all the information you need to solve [itex]\vec{x} \cdot \vec{y}[/itex], so you need to get the dot product you want to find in that form.

    12a) One possible way is to just use the distance formula to get the lengths of all three sides. Try Pythagoras's Theorem and see if it works out.

    However, this is a test on dot product, so whoever came up with the question probably wants you to use it. A fact you should know straight away is that the dot product of two orthogonal (means the same as perpendicular) vectors is 0. Why? Think about the formula for dot product and the angle between the two vectors.

    Plot the points, and treat the two sides that look like legs as vectors. How would you write the legs as vectors (what would the components of the "leg" vectors be if you moved their tails to the origin? Remember, two vectors can be equal even if their tails are in different locations). If the triangle was right, what would the dot product of the legs equal?

    12b) Well, if you found the length of the legs, that would be easy, since this is a right triangle. There is another way with cross products if you write two sides of the triangle as vectors. The area of a triangle given two sides [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] is

    [tex]A = \frac{1}{2}\cdot(\vec{u} \times \vec{v}) = \frac{1}{2}|\vec{u}||\vec{v}|\sin\theta[/tex]​

    If you know [itex]\theta[/itex], as you should in this case (if the two vectors are the legs), you can find the area. This is a direct consequence of a formula for the area of the triangle

    [tex]A = \frac{1}{2}ab\sin C[/tex]​

    where a and b are two sides of a triangle, and C is the angle between them.

    12c) ABS probably means ABC? Well, in this case, you it would be simplest just with the three sides. In fact, if you read this far, you may as well have just found the three sides from the beginning. Of course, if you've proved that the triangle is right using the dot product, you can find the third side with the lengths of the legs, found from the magnitudes of the vectors representing them.

    12d) Draw where you would like the fourth point to be. Move the vectors of the legs of the triangle around. Try adding them.

    13a) Projection onto one of the axes should be okay for you if you know how to project one vector onto another. You just need a vector in the direction of the axis you want--for simplicity, you can choose the unit vector.

    13b) Try drawing a diagram (I know you said it was hard for you, so practice). Plot the point and draw the vector--keep the tail at the origin for simplicity. Suppose you want to project the vector in the x-y plane (horizontal one). Drop a perpendicular from the head of the vector the the x-y plane, and connect where that perpendicular meets the plane to the origin. Isn't the projection you want just a vector from the origin to that intersection point? How would you find a vector for the perpendicular? How would you find the projected vector? Reply back if you still need help.

    14) See (12b).

    1) Talk yourself through these, or talk to us about them. Often for these questions, you should think about cosine and sine and the respective formulas using them for the dot and cross products. For 1a), you are confusing cross product with dot product. What should be the answer? Same for 1b).

    2) Right, the cross product will give you a vector orthogonal to both u and v. How would you make sure the length of this cross product vector is one (unit vector)?

    3) Make some tick marks on the x-axis also. Do your best to sketch. To get a better sketch, drop a dashed perpendicular to the x-y plane and lightly draw in the right triangle with the origin. This shows 3-D pretty well. Practice drawing in 3-D by looking at the textbook if necessary. Like you mentioned, see my response to 13. In any case, 3ai should be easy.
    Last edited: May 22, 2008
  12. May 22, 2008 #11
    #13. Find the projection of u=(17,-3,8)
    a) onto each of the coordinate axes
    b) onto each of the coordinate planes

    a) (17,0,0); (0,-3,0); ...

    b) not much sure .. but basically find magnitude and angle with each axis ..
    e.g. on x-y plane
    proj = magnitude * cos (angle that line makes with x-y plane)..
  13. May 22, 2008 #12
    Wow, typing all that was painful. I hope I didn't fudge anything up, but you've got second opinions on just about all of them from rootX and edziura :-) Good luck.
  14. May 23, 2008 #13
    Thanks everyone for your help. Especially Tedjn for all that typing!

    I'm still completely lost on 12b,12c and 14. What you typed doesn't even makes sense to me :confused:

    Anyways, at least i'm better off then I was before I posted here, even if I couldn't figure everything out. I attribute it to my current lack of sleep and the fact that it's 1AM here right now. :zzz:

    I'll read this thread again before my test, and we'll see what happens.

    I'll most likely be re-taking functions and calc. next year. I already got declined at both universities I applied at (for electrical engineering) even though I worked my tail off to get an 80% average. Last semester I got a 65% in chemistry, which killed my average, so if I re-take chem, functions, and calc, hopefully I can pull mid 80s in everything, and that should get me in September 09'.

    Anyways, G'nite :zzz:
  15. May 23, 2008 #14
    I'm going to sleep soon too, thank God for small favors :smile: I'll try to give a better explanation.

    12c) The best way to do this problem really is to just find the length of the three sides and add them. That gives you the perimeter. Just use the distance formula on the three points to find AB, AC, and BC.

    12b and 14) The idea about using the length of the legs of the right triangle still holds for 12b, but for 14, it would be easiest with cross products.

    You know that in general the area of a triangle is


    if a is the base and h is the height. From geometry, it can be shown that this is equivalent to

    [tex]\frac{1}{2}\,ab\sin C[/tex]​

    where a and b are two sides of the triangle, and C is the angle between them. See http://www.btinternet.com/~se16/hgb/triangle.htm and scroll down to the proofs. It's #2. Yes, there are a whole load of formulas for triangle area--that list doesn't even have them all.

    This bears striking resemblance to the equation of a cross product of two vectors:

    [tex]|\vec{u}\times\vec{v}| = |\vec{u}||\vec{v}|\sin\theta[/tex]​

    where [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] are two vectors, and [itex]\theta[/itex] the angle between them. When I say the angle between them, that is the angle you get when you put the two tails of the vectors together. Of course, there will be two angles, one is [itex]\theta[/itex] and one is [itex]360^\circ - \theta[/itex]. You should choose the smaller one.

    Let's take the triangle in #14 A(-7,3,0), B(3,1,0), and C(2,-6,0) and work the problem.

    Consider the vector [itex]\vec{a} = <-7, 3, 0>[/itex], [itex]\vec{b} = <3, 1, 0>[/itex], and [itex]\vec{c} = <2, -6, 0>[/itex]. These vectors start from the origin and end at the three vertices.

    Now, consider the vector [itex]\vec{AB} = \vec{b} - \vec{a}[/itex]. When you draw it out, this vector is just the side [itex]\overline{AB}[/itex] of the triangle with the tail at A and the head at B. Likewise, [itex]\vec{AC} = \vec{c} - \vec{a}[/itex] is just the same as side [itex]\overline{AC}[/itex] of the triangle, with the tail at A and the head at C. Convince yourself that the magnitude of [itex]\vec{AB}[/itex] is just the length AB, and the magnitude of [itex]\vec{AC}[/itex] is just the length AC. We now find that

    [tex]\vec{AB} = \vec{b} - \vec{a} = <3, 1, 0> - <-7, 3, 0> = <10, -2, 0>[/tex]​
    [tex]\vec{AC} = \vec{c} - \vec{a} = <2, -6, 0> - <-7, 3, 0> = <9, -9, 0>[/tex]​

    Label the angle between the two vectors [itex]\vec{AB}[/itex] and [itex]\vec{AC}[/itex] as [itex]\theta[/itex]. We see that:

    [tex]A = \frac{1}{2}(AB)(AC)\sin\theta = \frac{1}{2}|\vec{AB}||\vec{AC}|\sin\theta = \frac{1}{2}|\vec{AB} \times \vec{AC}|[/tex]​

    Of course, we don't know [itex]\theta[/itex]. We can either calculate it using the dot product, or we can just find the cross product using the determinant method:

    [tex]\vec{AB} \times \vec{AC} = \left|{\begin{array}{*{20}c} \hat{i} & \hat{j} & \hat{k} \\ 10 & -2 & 0 \\ 9 & -9 & 0 \\ \end{array}} \right| = -72\hat{k}[/tex]​


    [tex]A = \frac{1}{2}(72) = \fbox{36}[/tex]​

    Remember, the magnitude is always positive.

    Finally, this is not directed toward the test, but just to show you how there are sometimes many ways to solve a single problem and keep up your interest in math. Notice that none of the original three points A(-7,3,0), B(3,1,0), and C(2,-6,0) have a z-coordinate. Thus, all three points are entirely in the x-y plane, the points A(-7,3), B(3,1), and C(2, -6). From Algebra 2, you may have learned that the area of a triangle given 3 points in a plane is:

    [tex]A = \frac{1}{2}\left|{\begin{array}{*{20}c} {-7} & 3 & 1 \\ 3 & 1 & 1 \\ 2 & -6 & 1 \\ \end{array}}\right| = \frac{1}{2}[-49 + (-3) + (-20)] = -36[/tex]​

    We take the absolute value, so we once again find the that the area is 36. For more about this formula, see http://people.richland.edu/james/lecture/m116/matrices/applications.html in the middle.

    Anyway, good luck tomorrow.
    Last edited by a moderator: Apr 23, 2017
  16. May 23, 2008 #15
    Thanks again Tedjn.

    I totally forgot about the https://www.physicsforums.com/latex_images/17/1740837-11.png [Broken]

    My test is in about half an hour, i'm just reviewing my stuff now.
    Last edited by a moderator: May 3, 2017
  17. May 15, 2011 #16
    cos(inversecos(4/21)) = 4/21

    i know this is two years late and pretty useless but i have my test in a few days and I googled a question worded slightly different. Thanks, i know 7 now.

    17. The vectors x = (-4,p,-2) and y = (-2,3,6) are such that inversecos(4/21) = theta, where theta is the angle between x and y. Determine the value(s) of p.

    I ended up in the quadratic formula, I got a value of 4 and -0.68 approx, why is the negative value invalid?
  18. May 15, 2011 #17


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    Show a bit more of your work.
  19. May 15, 2011 #18


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    Try plugging it back into your original equation.
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