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Math  /  Geometry

QuestionSuppose a line is given parametrically by the equation L(t)=⟨1−4t,4−3t,4−t⟩L(t)=\langle 1-4 t, 4-3 t, 4-t\rangle
Then the vector and point that were used to define this line were vˉ=\bar{v}= □\square , and p=p= □\square (1,4,4)(1,4,4)

Studdy Solution

STEP 1

1. The line is given in parametric form.
2. The parametric equation of the line is L(t)=⟨1−4t,4−3t,4−t⟩ L(t) = \langle 1-4t, 4-3t, 4-t \rangle .

STEP 2

1. Identify the vector used in the parametric equation.
2. Identify the point used in the parametric equation.

STEP 3

Identify the vector used in the parametric equation:
The vector vˉ\bar{v} is the coefficient of tt in the parametric equation. From L(t)=⟨1−4t,4−3t,4−t⟩L(t) = \langle 1-4t, 4-3t, 4-t \rangle, the vector vˉ\bar{v} is:
vˉ=⟨−4,−3,−1⟩ \bar{v} = \langle -4, -3, -1 \rangle

STEP 4

Identify the point used in the parametric equation:
The point pp is the constant part of the parametric equation when t=0t = 0. From L(t)=⟨1−4t,4−3t,4−t⟩L(t) = \langle 1-4t, 4-3t, 4-t \rangle, the point pp is:
p=⟨1,4,4⟩ p = \langle 1, 4, 4 \rangle
The vector and point that were used to define this line are:
vˉ=⟨−4,−3,−1⟩ \bar{v} = \langle -4, -3, -1 \rangle p=⟨1,4,4⟩ p = \langle 1, 4, 4 \rangle

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