Question

Point Parallel to {−6,0,π)∀−84+2}−k=k (a) narametre equations (Enter your arowers as a cormma- ieporated list.) (b) symmetric equations 8x−y=628x=2y=6z8x+6=2y=6y−z8x−6=y=6z

Answer 1

To write the equations of a line in parameter and symmetric form, use the direction vector and a point on the line.

Step-by-step explanation:

To write the equations in parameter form, we can use the form: x = x0 + at, y = y0 + bt, z = z0 + ct, where (a, b, c) is a direction vector parallel to the line and (x0, y0, z0) is a point on the line. In this case, the direction vector is (-8, 2, 6) and a point on the line is (-6, 0, π). So, the parameter equations become: x = -6 - 8t, y = 0 + 2t, z = π + 6t.

To write the symmetric equations, we set each of the parameter equations equal to a constant and solve for t. x = -6 - 8t -> t = (-x - 6)/8. y = 0 + 2t -> t = (y - 0)/2. z = π + 6t -> t = (z - π)/6. Substituting these values of t into the respective parameter equations, we get: (-x - 6)/8 = (y - 0)/2 = (z - π)/6.

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