Question

Find a vector equation and parametric equation for the line segment that joins P to Q. P(-7,6,0) Q (3, -1,1,5)

(a) Vector equation R(t) = ( 5t)
(b) Parametric equations.

Answer 1

The vector equation for the line segment joining P(-7, 6, 0) to Q(3, -1, 5) is R(t) = (-7, 6, 0) + t(10, -7, 5), and the parametric equations are x(t) = -7 + 10t, y(t) = 6 - 7t, and z(t) = 5t.

Step-by-step explanation:

To find the vector equation and parametric equations for the line segment that joins points P and Q, we need to first identify the coordinates of P and Q. Based on the provided information, point P is at (-7, 6, 0) and Q seems to have a typographical error. Assuming Q should be a 3-dimensional point, let's correct it to (3, -1, 5). We find the direction vector by subtracting the coordinates of P from Q: (3-(-7), -1-6, 5-0) = (10, -7, 5).

The vector equation for the line is given by:

R(t) = P + tD where D is the direction vector and t is a parameter. Substituting in the values gives:

R(t) = (-7, 6, 0) + t(10, -7, 5)

This can also be written in parametric form as:

x(t) = -7 + 10t
y(t) = 6 - 7t
z(t) = 5t