Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −8) and parallel t

Question

Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −8) and parallel to the line x = −1 + 4t, y = 6 − 4t, z = 3 + 6t

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Rose 2 weeks 2021-09-10T09:58:26+00:00 1 Answer 0

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    2021-09-10T10:00:14+00:00

    Answer:

    The vector equation of the line is \overrightarrow {r}=(11j-8k)+t(4i-4j+6k) and parametric equations for the line are x=4t, y=11-4t, z=-8+6t.

    Step-by-step explanation:

    It is given that the line passes through the point (0,11,-8) and parallel to the line

    x=-1+4t

    y=6-4t

    z=3+6t

    The parametric equation are defined as

    x=x_1+at,y=y_1+bt,z=z_1+ct

    Where, (x₁,y₁,z₁) is point from which line passes through and <a,b,c> is cosine of parallel vector.

    From the given parametric equation it is clear that the line passes through the point (-1,6,3) and parallel vector is <4,-4,6>.

    The required line is passes through the point (0,11,-8) and parallel vector is <4,-4,6>. So, the parametric equations for the line are

    x=4t

    y=11-4t

    z=-8+6t

    Vector equation of a line is

    \overrightarrow {r}=\overrightarrow {r_0}+t\overrightarrow {v}

    where, \overrightarrow {r_0} is a position vector and \overrightarrow {v} is cosine of parallel vector.

    \overrightarrow {r}=(11j-8k)+t(4i-4j+6k)

    Therefore the vector equation of the line is \overrightarrow {r}=(11j-8k)+t(4i-4j+6k) and parametric equations for the line are x=4t, y=11-4t, z=-8+6t.

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