Solve Coordinate Geometry Problem: Find Line Equation

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To find the equation of a line in three dimensions, it is essential to use parametric equations rather than a single equation, which can mistakenly represent a plane. The line must pass through the point (3, 4, -1) and be parallel to the vector 2i - 3j + 6k. The correct parametric equations are x = 3 + 2t, y = 4 - 3t, and z = -1 + 6t, where t is a parameter. This approach clarifies the relationship between the line and the direction vector. Understanding these concepts is crucial for solving coordinate geometry problems effectively.
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I think my coordinate geometry skills have turned rather rusty. :frown:
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 :bugeye: . I am all confused now. How do I relate this line to the vector parallel to it?
 
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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.
 

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