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 , , .

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation:

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: