Question

Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −8) and parallel to the line x = −1 + 4t, y = 6 − 4t, z = 3 + 6t

Answer 1

Answer:

The vector equation of the line is $\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)$ and parametric equations for the line are $x=4t$, $y=11-4t$, $z=-8+6t$.

Step-by-step explanation:

It is given that the line passes through the point (0,11,-8) and parallel to the line

$x=-1+4t$

$y=6-4t$

$z=3+6t$

The parametric equation are defined as

$x=x_1+at,y=y_1+bt,z=z_1+ct$

Where, (x₁,y₁,z₁) is point from which line passes through and <a,b,c> is cosine of parallel vector.

From the given parametric equation it is clear that the line passes through the point (-1,6,3) and parallel vector is <4,-4,6>.

The required line is passes through the point (0,11,-8) and parallel vector is <4,-4,6>. So, the parametric equations for the line are

$x=4t$

$y=11-4t$

$z=-8+6t$

Vector equation of a line is

$\overrightarrow {r}=\overrightarrow {r_0}+t\overrightarrow {v}$

where, $\overrightarrow {r_0}$ is a position vector and $\overrightarrow {v}$ is cosine of parallel vector.

$\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)$

Therefore the vector equation of the line is $\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)$ and parametric equations for the line are $x=4t$, $y=11-4t$, $z=-8+6t$.