Question

Give a vector parametric equation for the line through the point (−2,−4,0) that is parallel to the line ⟨−2−3t,1−4t,−5t⟩:

Answer 1

The tangent vector for the given line is

T(t) = d/dt ⟨-2 - 3t, 1 - 4t, -5t⟩ = ⟨-3, -4, -5⟩

On its own, this vector points to a single point in space, (-3, -4, -5).

Multiply this vector by some scalar t to get a whole set of vectors, essentially stretching or contracting the vector ⟨-3, -4, -5⟩. This set is a line through the origin.

Now translate this set of vectors by adding to it the vector ⟨-2, -4, 0⟩, which correspond to the given point.

Then the equation for this new line is simply

L(t) = ⟨-3, -4, -5⟩t + ⟨-2, -4, 0⟩ = ⟨-2 - 3t, -4 - 4t, -5t⟩

Answer 2

The vector parametric equation for the line through the point is $r = (-2, \ -4, \ 0) +(-3, \ -4, \ -5)t$.

Given

Give a vector parametric equation for the line through the point (−2,−4,0) that is parallel to the line ⟨−2−3t,1−4t,−5t⟩.

What is a parametric equation vector?

Parametric equations of the line segment are defined by its endpoints.

To find the vector equation of the line segment, we’ll convert its endpoints to their vector equivalents.

Two lines are parallel if they have the same direction, and in the parametric form, the direction of a line is always the vector of constants that multiply t (or the parameter).

The vector equation of a line is given by:

$\rm v = r_0+tv$

Where v is the direction vector and $\rm r_0$ is a point of the line.

The tangent vector for the given line is

T(t) = d/dt ⟨-2 - 3t, 1 - 4t, -5t⟩ = ⟨-3, -4, -5⟩

Here,

$\rm r_0 = (-2,-4,0) \ and \ v=(-3, \ -4, \ -5)t$

Then,

$r = (-2, \ -4, \ 0) +(-3, \ -4, \ -5)t$

x = -2-3t, y = -4-4t, and z = 0-5t

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