Final answer:
To find the parametric and symmetric equations of the line through the points (3,1,-1) and (3,2,-6), calculate the direction vector (0,1,-5). Parametric Form: x = 3, y = 1 + t, z = -1 - 5t. The symmetric equations cannot include x since the line is parallel to the yz-plane, but we have \((y-1)/1 = (z+1)/-5\).
Step-by-step explanation:
To find the parametric equations and symmetric equations for the line through the points (3,1,-1) and (3,2,-6), we must first determine the direction vector of the line by subtracting the corresponding coordinates of the points, which gives us (0,1,-5). This direction vector tells us how to move from one point to another along the line.
Parametric Equations:
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- x = 3
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- y = 1 + t
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- z = -1 - 5t
Symmetric Equations:
Since the x-values of our points are the same, the line is parallel to the yz-plane. Therefore, the x-component does not change, which is why there is no symmetric equation involving x. However, the y and z components give us:
\(\frac{y-1}{1} = \frac{z+1}{-5}\)