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Math  /  Geometry

QuestionGive a vector parametric equation for the line through the point (−1,0,4)(-1,0,4) that is parallel to the line ⟨4−5t,4−4t,5−2t⟩\langle 4-5 t, 4-4 t, 5-2 t\rangle : L(t)=L(t)=

Studdy Solution

STEP 1

1. The line is defined in three-dimensional space.
2. The given line is expressed in parametric form.
3. The new line must pass through the point (−1,0,4)(-1, 0, 4).
4. The new line must be parallel to the given line.

STEP 2

1. Identify the direction vector of the given line.
2. Use the direction vector and the given point to write the parametric equation of the new line.

STEP 3

Identify the direction vector of the given line:
The given line is ⟨4−5t,4−4t,5−2t⟩\langle 4-5t, 4-4t, 5-2t \rangle. The direction vector can be found by examining the coefficients of tt in each component:
Direction vector = ⟨−5,−4,−2⟩\langle -5, -4, -2 \rangle.

STEP 4

Use the direction vector and the given point to write the parametric equation of the new line:
The parametric equation of a line through a point (x0,y0,z0)(x_0, y_0, z_0) with direction vector ⟨a,b,c⟩\langle a, b, c \rangle is given by:
r(t)=⟨x0,y0,z0⟩+t⋅⟨a,b,c⟩ \mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t \cdot \langle a, b, c \rangle
Substitute the given point (−1,0,4)(-1, 0, 4) and the direction vector ⟨−5,−4,−2⟩\langle -5, -4, -2 \rangle:
r(t)=⟨−1,0,4⟩+t⋅⟨−5,−4,−2⟩ \mathbf{r}(t) = \langle -1, 0, 4 \rangle + t \cdot \langle -5, -4, -2 \rangle
r(t)=⟨−1−5t,0−4t,4−2t⟩ \mathbf{r}(t) = \langle -1 - 5t, 0 - 4t, 4 - 2t \rangle
The vector parametric equation for the line is:
L(t)=⟨−1−5t,−4t,4−2t⟩ L(t) = \langle -1 - 5t, -4t, 4 - 2t \rangle

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