(1 point) Suppose a line is given parametrically by the equation L(t)=⟨−1−4t,−t,3−t⟩L(t)=⟨−1−4t,−t,3−t⟩ Then the vector and point that were used to define this line were v¯¯¯v¯ = , and pp = .

If we want to find the director vector and the point that were used to define this line we need to rewrite the equation of the line in the vectorial equation.

We can write (using factoring) :

[tex](-1-4t,-t,3-t)=(-1,0,3)+t(-4,-1,-1)[/tex]

Therefore, we can write the line in the vectorial form as :

[tex]L:\left[\begin{array}{c}x&y&z\\\end{array}\right]=\left[\begin{array}{c}-1&0&3\\\end{array}\right]+t\left[\begin{array}{c}-4&-1&-1\\\end{array}\right][/tex] with t ∈ |R

The director vector of the line is the vector which is multiplying the parameter ( ''t'' in this case). Therefore the director vector is

[tex]v=\left[\begin{array}{c}-4&-1&-1\\\end{array}\right][/tex]

The point that was used to define this line is the another vector in the vectorial equation of the line. In this case :

[tex]p=\left[\begin{array}{c}-1&0&3\\\end{array}\right][/tex]