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Homework Help: Whether lines are || or perendicular

  1. Jun 11, 2010 #1
    1. The problem statement, all variables and given/known data

    State whether lines are || or perpendicular, and find the angle between the lines

    line A is given by the following parametric equations:

    x = 1 -3t
    y = 2 +4t
    z = -6 +t

    line B:

    x = 1 +2s
    y = 2 -2s
    z = -6 +s

    2. Relevant equations

    3. The attempt at a solution

    I looked at both z values -6+t = -6+s this is only true if and only if s = t for any value.
    Neverthless, I started with the dumb way, setting each equation equal to the other.

    1 +2s = 1 -3t and we found s = -3/2 and substitute this back to the y equations.
    2+4t = 2-2(-3/2) => t = 3/4
    this clearly showed that s =/= t but and z cannot be equal...

    i looked at the book's answer, it gave me a cos-1 (-13/sqrt(234))

    but only if lines are perpendicular there is an angle delta between the lines. neither skew nor || lines !!!! so why did the book answer the second question? there shouldn't even have an angle between the two planes.
  2. jcsd
  3. Jun 11, 2010 #2


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    Homework Helper

    Write both lines in the form r= a +tu

    then find the dot product of the directions.

    If the dot product is zero they are perpendicular, if the cross-product is zero, they are parallel.
  4. Jun 11, 2010 #3
    right. i totally ignored the dot product angle. thanks
    and of course, the two lines is neither || nor orthogonal

    but how do i know when should i use the cross or dot product angle?
  5. Jun 11, 2010 #4
    I think you are misusing the word "angle". The cross product gives you another vector; the dot product gives you access to angles.
  6. Jun 11, 2010 #5
    Hi. Thank you for point it out.
    Yes. I probably got confused with this

    For A cross B, if delta is the angle between a and b, then
    |\mathbf{A}\times\mathbf{B}| = AB \sin \theta
    The above defintion reminds me of plane u x v = nornmal vector, and this is how i think of the angle between the two vectors (as a plane).

    However, the dot product cos delta = a*b /|a| |b| is defines as:
    Let delta be the angle between nonzero vectors a and b, such that a * b = |a| |b| cos delta

    I am confused with the two definitions. They both mentioned "vectors A and B, although their properties are different (cross vs dot), both said something about "delta is the angle between vector a and b".
  7. Jun 11, 2010 #6


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    Dot product gives a scalar

    Cross product gives another vector.

    If you want the angle between two vector that lie in the same plane, you can use either really.
  8. Jun 11, 2010 #7
    Hi rock. Thank you. Right.

    I did the calculation:

    for dot product a*b / |a| |b| = cos-1 i got -13/sqrt(26) *3 which is exactly what the books give. (angle => 31.8 degree

    for the cross product |a cross b| / |a| |b| i got sqrt(65) / sqrt(26)*3 and this is does not return the same angle after conversion. this gives 148.2 degree

    how can i verify? or did i misinterpret your statement?
  9. Jun 11, 2010 #8


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    They would give the same answer.

    [tex]sin^{-1}(\frac{\sqrt{65}}{3\sqrt{26}}) = 31.8[/tex]

    They give the same acute angle.

    But it is better to use the dot product since you do not know whether the angle is acute or obtuse using the cross-product.

    if a.b is negative, your angle is obtuse.
  10. Jun 11, 2010 #9
    Thank you. I forgot about that too.
  11. Jun 12, 2010 #10


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    None of these responses actually answer jwxie's initial question- how can you say two lines intersect at a given angle if they are skew? Even if the dot product of two vectors is 0, the lines having those direction vectors are not "perpendicular" if they do not intersect.

    Answer- they are not skew. Yes, from the z equations, we get s= t. Putting that into y= 2+ 4t= 2- 2s you get 2+ 4t= 2- 2t so that 4t= -2t and t= s= 0. Then the x equations result in x= 1. The lines intersect at (1, 2, -6).
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