Angle between Vectors

Key Questions

  • When we are to consider the angle between any two vectors, it should be noted that the angle which is less than #pi# is to be taken.

  • You can use the dot product to solve this problem. See http://en.wikipedia.org/wiki/Dot_product

    The dot product is an operation on two vectors. There are two different definitions of dot product. Let #\vec(A)=[A_1,A_2,...,A_n]# be a vector and #\vec(B)=[B_1,B_2,...,B_n]# be another vector, then we have 2 formulas for dot product:

    1) Algebraic definition:

    #\vec(A) \cdot \vec(B) = \sum_1^n A_i B_i = A_1 B_1 + A_2 B_2 + ... + A_n B_n#

    2) Geometric definition:

    #\vec(A) \cdot \vec(B) = ||\vec(A)||\ ||\vec(B)||\cos(\theta)#

    where #\theta# is the angle between #\vec(A)# and #\vec(B)#, and #||\vec(A)||# denotes the magnitude of #\vec(A)# and has the formula:

    #||\vec(A)|| = \sqrt(A_1^2 + A_2^2 + ... + A_n^2)#

    We can solve many questions (such as the angle between two vectors) by combining the two definitions:

    #\sum_1^n A_i B_i = ||\vec(A)||\ ||\vec(B)||\cos(\theta)#

    or

    #A_1 B_1 + A_2 B_2 + ... + A_n B_n = (\sqrt(A_1^2 + A_2^2 + ... + A_n^2))(\sqrt(B_1^2 + B_2^2 + ... + B_n^2))\cos(\theta)#

    If we have two vectors, then the only unknown is #\theta# in the above equation, and thus we can solve for #\theta#, which is the angle between the two vectors.

    Example:

    Q: Given #\vec(A) = [2, 5, 1]#, #\vec(B) = [9, -3, 6]#, find the angle between them.

    A:
    From the question, we see that each vector has three dimensions. From above, our formula becomes:

    #A_1 B_1 + A_2 B_2 + A_3 B_3 = (\sqrt(A_1^2 + A_2^2 + A_3^2))(\sqrt(B_1^2 + B_2^2 + B_3^2))\cos(\theta)#

    Left side:

    #A_1 B_1 + A_2 B_2 + A_3 B_3 = (2)(9) + (5)(-3) + (1)(6) = 9#

    Right side:

    #||\vec(A)|| = \sqrt(A_1^2 + A_2^2 + A_3^2) = \sqrt(2^2 + 5^2 + 1^2) = \sqrt(30)#
    #||\vec(B)|| = \sqrt(B_1^2 + B_2^2 + B_3^2) = \sqrt(9^2 + (-3)^2 + 6^2) = \sqrt(126)#
    #\theta# is unknown

    Plug everything into the formula, we get:

    #9 = (\sqrt(30))(\sqrt(126))\cos(\theta)#

    Solve for #\theta#:

    #\cos(\theta) = \frac(9)((\sqrt(30))(\sqrt(126))#
    #\theta = \cos^-1(\frac(9)((\sqrt(30))(\sqrt(126))))#

    Using a calculator, we get:

    #\theta = 81.58# degrees

    See the following video of ...

    Angle Between Vectors Example

  • It is simply the product of the modules of the two vectors (with positive or negative sign depending upon the relative orientation of the vectors).
    A typical example of this situation is when you evaluate the WORK done by a force #vecF# during a displacement #vecs#.
    For example, if you have:
    enter image source here
    Work done by force #vecF#:
    #W=|vecF|*|vecs|*cos(theta)#
    Where #theta# is the angle between force and displacement; the two vectors being parallel can give:

    #theta=0°# and #cos(theta)=cos(0°)=1# so:
    #W=5*10*1=50 J#

    Or:

    #theta=180°# and #cos(theta)=cos(180°)=-1# so:
    #W=5*10*-1=-50 J#

  • The dot of two vectors is given by the sum of its correspondent coordinates multiplied. In mathematical notation:
    let #v = [v_(1), v_(2), ... , v_(n)] # and #u = [u_(1), u_(2), ... , u_(n)]#,
    Dot product:
    #v*u = #
    #sum v_(i).u_(i) = (v_(1).u_(1)) + (v_(2).u_(2)) + ... + (v_(n).u_(n))#

    and angle between vectors:
    #cos(theta) =(v*u)/(|v||u|)#

    Since the angle between two perpendicular vectors is #pi/2#, and it's cosine equals 0:
    #(v*u)/(|v||u|) = 0 :. v*u = 0#

    Hope it helps.

Questions