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A line passes through the point (3 comma 1 comma negative 8 )​, and is parallel to the vector 5 Bold i plus 3 Bold j plus 3 Bold k. Find the standard parametric equations for the​ line, written using the components of the given vector and the coordinates of the given point.

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LammettHash

The line [tex]\ell[/tex] passing through (3, 1, -8) and running parallel to the vector <5, 3, 3> has the vector equation

[tex]\ell=(3\,\vec\imath+\vec\jmath-8\,\vec k)+t(5\,\vec\imath+3\,\vec\jmath+3\,\vec k)[/tex]

or

[tex]\ell=(3+5t)\,\vec\imath+(1+3t)\,\vec\jmath+(3t-8)\,\vec k[/tex]

and from this we can extract a set of parametric equations for the line:

[tex]\begin{cases}x(t)=5t+3\\y(t)=3t+1\\z(t)=3t-8\end{cases}[/tex]

where [tex]t\in\Bbb R[/tex].

The parametric equations of the line are given by:

[tex]x(t) = 3 + 5t[/tex]

[tex]y(t) = -1 + 3t[/tex]

[tex]z(t) = -8 + 3t[/tex]

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A line passing through the point [tex](x_0, y_0, z_0)[/tex] and being parallel to a vector ai + bj + ck has parametric equations given by:

[tex]x(t) = x_0 + at[/tex]

[tex]y(t) = y_0 + bt[/tex]

[tex]z(t) = z_0 + ct[/tex]

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  • Point (3, -1, -8) means that [tex]x_0 = 3, y_0 = -1, z_0 = -8[/tex].
  • Parallel to vector 5i + 3j + 3k means that [tex]a = 5, b = 3, c = 3[/tex]

Thus

[tex]x(t) = 3 + 5t[/tex]

[tex]y(t) = -1 + 3t[/tex]

[tex]z(t) = -8 + 3t[/tex]

A similar problem is given at https://brainly.com/question/15986055

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