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

The vector equation of the line is and parametric equations for the line are , , .

Step-by-step explanation:

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

The parametric equation are defined as

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

Vector equation of a line is

where, is a position vector and is cosine of parallel vector.

Therefore the vector equation of the line is and parametric equations for the line are , , .

7

step-by-step explanation:

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64