slader (a) Find parametric equations for the line through (4, 1, 8) that is perpendicular to the plane x − y + 4z = 2. (Use the parameter t.) (x(t), y(t), z(t)) = Correct: Your answer is correct. (b) "In what points does this line intersect the coordinate planes?"

(x(t), y(t), z(t)) = (4 + t, 1 - t, 8 + 4t)

xy - plane    (x, y, z) = (2, -1, 0)

yz - plane    (x, y, z) = (0, 5, -8)

xz - plane     (x, y, z) = (5, 0, 12)

Step-by-step explanation:

The given point (x, y ,z) = (4, 1, 8)

The plane x -y + 4z = 2

Normal vector (n) = < 1, -1, 4 >

The equation of line through point (4, 1, 8) and the plane is:

(x(t), y(t), z(t)) = (4, 1, 8) + t(1, -1, 4)

(x(t), y(t), z(t)) = (4 + t, 1 - t, 8 + 4t)

Any point on the line P(x, y, z) = ( 4 + t, 1 - t, 8 + 4t)

xy-Pane ⇒ z = 0

8 + 4t = 0

4t = - 8

t = -8/4

t = -2

∴

(x, y, z) = (4 - 2, 1 - 2, 8 + 4(-2))

(x, y, z) = (2, -1, 0)

yz-plane ⇒ x = 0

4 + t = 0

t = -4

∴

(x, y, z) = (4 + (-4) , 1-(-4), 8 + 4(-4)

(x, y, z) = (0, 5, -8)

xz-plane ⇒ y = 0

1 - t = 0

-t = -1

t = 1

∴

(x, y, z) = ( 4 + 1, 1 - 1, 8 + 4(1) )

(x, y, z) = (5, 0, 12)