Saturday
March 25, 2017

Post a New Question

Posted by on .

Given two non-collinear vectors a and b, show that a, axb, and (axb)xa are mutually perpendicular.

Also

Express the unit vectors i, j, and k as ordered triplets and show that
i x j = k
what are ordered triplets?


Using components show that
u x v = -v x u for any vectors u and v.

  • Math - ,

    A cross product is always perpendicular to BOTH vectors, unless it the cross product is zero, which happens only for collinear vectors.

    Therefore (a x b) is perpendicular to a.

    My statement also proves that (axb)x a is perpendicular to both a and a x b

    The "proof" they want you to do involves actually computing the cross product of a x b where a = i and b = j. You will end up proving that the cross product is k. Use these "ordered triplet" designations:
    i = (1 0 0)
    j = (0 1 0)
    k = (0 0 1)

    i x j = value of the determinant
    |i j k|
    |1 0 0|
    |0 1 0| = 0i + 0j + k

  • Math - ,

    Your last question <<Using components show that u x v = -v x u for any vectors u and v.>>
    can be done similarly using arbitrary ordered triplet designations for u and v,
    u = ai + bj + ck
    v = di + ej + fk
    where a, b, c, d, e and f are arbitrary constants and then using the determinant formula to compute the cross products. u x v and v x u

    u x v =
    |i j k|
    |a b c|
    |d e f|
    = (bf - ce)i + (cd - af)j + (ae - bd)k
    Now compare that to v x u =
    |i j k|
    |d e f|
    |a b c| = (ec - bf)i, etc

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question