Solve Coordinate Geometry Problem: Find Line Equation

[Reshma]

I think my coordinate geometry skills have turned rather rusty.
Just help me out with this problem.

Find the symmetric and parametric equation of a line passing through (3, 4, -1) and parallel to 2i - 3j + 6k.

Here is my attempt at this:

Since the line passes through (3,4,-1) the equation would be given by:
(x - 3) + (y - 4) + (z +1) = 0

But, this looks like the equation of a plane . I am all confused now. How do I relate this line to the vector parallel to it?

 

[HallsofIvy]

Yes, that's the equation of the plane through (3, 4, -1) and perpendicular to 2i- 3j+ 6k. You can't write a line in 3 dimensions in just one equation. Write it in parametric equations:
x= x₀+ at, y= y₀+ bt, z= z₀+ ct where (x₀, y₀, z₀) is a point on the line and ai+ bj+ ck is a vector in the direction of the line.

 

[Architeuthis Dux]

Hello there! Don't worry, I can help you with this coordinate geometry problem. First, let's review some basics. A line in 3-dimensional space can be represented by the general equation:
(x - x1)/a = (y - y1)/b = (z - z1)/c
where (x1, y1, z1) is a point on the line and (a, b, c) is a direction vector parallel to the line. In your problem, the given vector 2i - 3j + 6k is already in the form of (a, b, c). So, we can use it to find the line equation.

To find the symmetric equation, we can simply plug in the given point (3, 4, -1) and the direction vector (2, -3, 6) in the general equation. This will give us:
(x - 3)/2 = (y - 4)/-3 = (z + 1)/6

To find the parametric equation, we can use the following formula:
x = x1 + at
y = y1 + bt
z = z1 + ct
where t is a parameter. Again, plugging in the given point and direction vector, we get:
x = 3 + 2t
y = 4 - 3t
z = -1 + 6t

So, the symmetric equation of the line is (x - 3)/2 = (y - 4)/-3 = (z + 1)/6 and the parametric equation is x = 3 + 2t, y = 4 - 3t, z = -1 + 6t. I hope this helps you understand and solve the problem. Keep practicing and your coordinate geometry skills will be back in no time!