Find Equation of Line Passing Through $(1, -2, -3)$ & Perpendicular to Plane

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SUMMARY

The equation of the line passing through the point $(1, -2, -3)$ and perpendicular to the plane defined by the equation $3x - y - 2z + 4 = 0$ is correctly formulated as $\overrightarrow{l}(t) = (1, -2, -3) + t \overrightarrow{v}$, where the perpendicular vector of the plane is $\overrightarrow{v} = (3, -1, -2)$. This formulation effectively demonstrates that the line is parallel to the normal vector of the plane, confirming its perpendicularity. The approach taken is efficient and accurate for deriving the line's equation.

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  • Basic concepts of perpendicularity in vector geometry
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mathmari
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Hello! :o

Find the equation of the line that passes through $(1, -2, -3)$ and that is perpendicular to the plane $3x-y-2z+4=0$.

Can we formulate it as followed??

The perpendicular vector of the plane is $\overrightarrow{v}=(3, -1, -2)$.

Since the line that we are looking for is also perpendicular to the plane, we have that the line is parallel to the vector $\overrightarrow{v}$.

Therefore, the line is $$\overrightarrow{l}(t)=(1, -2, -3)+t \overrightarrow{v}$$

Is it correct?? (Wondering)

Could I improve something at the formulation ?? (Wondering)
 
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Hi,

It's OK, and I think is the "cheapest" way to find the equation, good work! :cool:
 
Fallen Angel said:
Hi,

It's OK, and I think is the "cheapest" way to find the equation, good work! :cool:

Great! Thank you very much! (Smile)
 

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