CALC III: Vectors problem?

Let a = < 5, 3, -4 > and b = <1, 0, 2>. Show that there are scalars s and t so that sa + tb = < -13, -9, 16 >

Find

s=

t=

2 Answers

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  • John
    Lv 4
    7 years ago

    Plug in the vectors a and b

    s<5, 3, -4> + t<1, 0, 2> = <-13, -9, 16>

    Add the components and equate them to the components of the last vector

    5s + t = -13

    3s = -9

    -4s + 2t = 16

    Solve for s and t, use the 2nd equation, s = -3. Plug this into the first equation:

    (-15) + t = -13, solve for t = 2

    So, plug in the scalars s = -3 and t = 2 for the last equation. If the equation holds true, then there are scalars s and t for which sa + tb = <-13, -9, 16>

    Plug in: -4(-3) + 2(2) = 12 + 4 = 16

    This holds true as 16 is the last component of the vector <-13, -9, 16>.

    Hope this helps.

  • 7 years ago

    a = 5i + 3j -4k

    b = i + 2k

    if you simply added these vectors together you would get 6i + 3j - 2k

    So obviously there are scalars s & t that will give the answer of -13i - 9j + 16k

    s(5i + 3j - 4k) + t(i + 2k) = -13i - 9j + 16k

    solve for the scalars

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