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Math Help - Finding a component of a coplanar vector of two other vectors

  1. #1
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    Finding a component of a coplanar vector of two other vectors

    The question is determine the value of x for which the vector c = xi + j + k is coplanar with the vectors a ( i +j -k) and B (2i - j + 3k).
    The anser is x = 4 and i think u need to use vector triple cross product to find it but i cant seem to manage it or find any examples of this style of question. Its seems to be a common style of question on most of the past papers of which my exam is tomoz.

    Thanks in advance John
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  2. #2
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    Hello,
    Quote Originally Posted by mr_sputnik View Post
    The question is determine the value of x for which the vector c = xi + j + k is coplanar with the vectors a ( i +j -k) and B (2i - j + 3k).
    The anser is x = 4 and i think u need to use vector triple cross product to find it but i cant seem to manage it or find any examples of this style of question.

    Thanks in advance John
    Since vectors a and b are not colinear, they define a plane.

    If you make the cross product between them two, you will get a vector d perpendicular to them two, that is, by extension, a vector orthogonal to the plane they form.

    Thus, once you have the coordinates of d, find vector c such that its scalar product with d is 0, because any vector in this plane will be orthogonal to d.
    Since you already have 2 coordinates of c, it will be easy
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  3. #3
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    ur an ace chears, got correct anser
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  4. #4
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    Hello, John!

    Determine the value of x for which the vector \vec{C} \:= \:xi + j + k
    is coplanar with the vectors: \vec{A} \:=\: i +j -k and \vec{B} \:=\:2i - j + 3k

    The answer is x = 4
    I think u need to use vector triple scalar product.

    Recall that the triple scalar product, \vec{C} \cdot (\vec{A} \times \vec{B}), produces the volume
    . . of the parallelepiped ('slanted box") determined by the three vectors.

    If the three vectors are coplanar, the volume of the "box" is, of course, zero.

    So we have: . C\cdot(A \times B) \;=\;\left|\begin{array}{ccc}x & 1 & 1 \\ 1 & 1 & \text{-}1 \\ 2 & \text{-}1 & 3 \end{array}\right| \;=\;x(3-1)-(3+2) + (\text{-}1-2) \;=\;2x-8


    Therefore: . 2x - 8 \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:4}

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  5. #5
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    if u read what i put more carefully rather than quoting me wroung you would see i put vector triple product ( a x (b x c) )with is differnt from the scalar triple product (a . (B x C) that is used to vol of parralapipehead.
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  6. #6
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    Hello, mr_sputnik!

    if u read what i put more carefully rather than quoting me wroung you would see i put vector triple product ( a x (b x c) )with is differnt from the scalar triple product (a . (B x C) that is used to vol of parralapipehead.

    I did read what you wrote very carefully. . I did the problem correctly,
    . . gave you the proper procedure and the correct answer.

    So save your scolding for someone who did something stupid, okay?



    Since your triple cross product leads nowhere,

    . . I assumed that you meant triple scalar product

    . . because (if you read my work) that is the proper formula for the problem.



    Tale a look at your vector triple cross product . . .

    A \times (B \times C) \;=\;\langle x,1,1\rangle \times \left|\begin{array}{ccc} i & j & k \\ 1 & 1 & \text{-}1 \\ 2 & \text{-}1 & 3 \end{array}\right| \;=\;\langle x,1,1\rangle  \times \langle 2,-5,3\rangle

    . . . . . . = \;\left|\begin{array}{ccc}i & j & k \\ x & 1 & 1 \\ 2 & \text{-}5 & \text{-}3 \end{array}\right| \;=\;\langle 2,\:3x+2,\:-5x-2\rangle


    So you have a vector . . . now what?

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