Question

. Find sets of parametric equations and symmetric equations of the line through the point parallel to the given vector or line (if possible).
Point (-4,0,2)


Parallel to v=2i + 8j - 7k


(a) parametric equations (Enter your answers as a comma-separated list.)


(b) symmetric equations


A. 2x= y/8 = 7z


B. (x+4)/2 = y/8 = (2-z)/7


C x/2 = y = z/7


D. (x-4)/2 = y = z/7

Answer 1

Answer:

a) L(x,y,z) = (-4,0,2)+(2,8,-7)*t

b) (2-z)/7= y/8=(x+4)/2 (option B)

Step-by-step explanation:

the parametric equation of the line passing through the point P₀= (-4,0,2) and parallel to the vector v=2i + 8j - 7k is

L(x,y,z)=P₀+v*t

therefore

L(x,y,z) = (-4,0,2)+(2,8,-7)*t

or

x=x₀+vx*t = -4 + 2*t

y=y₀+vy*t = 8*t

z=z₀+vz*t = 2 -7*t

solving for t in the 3 equations we get the symmetric equation of the line:

(2-z)/7= y/8=(x+4)/2

thus the option B is correct

Answer 2

Answer:

a) Parametric equations

x = -4 + 2t

y = 8t

z = 2-7t

b) symmetric equations

$\frac{x+4}{2}= \frac{y}{8} = \frac{z+2}{-7}$          The answer is the option B

Step-by-step explanation:

For writing the vectorial equation of a line, we need a point in the line and its director vector, thus:

$L: (x_{0},y_{0},z_{0} ) + t(a,b,c)$

                            Where   $(x_{0},y_{0},z_{0})$ is a point in the line

                                           (a,b,c) is the director vector

Then

$L: (-4,0,2) + t(2,8,-7)$

a) Parametric equations

Since the vectorial equation, we can obtain the parametric equations writing the equation for each component

x = -4 + 2t

y = 8t

z = 2-7t

b)Symmetric equations

Since the parametric equations, we isolate the parameter t

$\frac{x+4}{2}= \frac{y}{8} = \frac{z+2}{-7}$