- #1
mathmari
Gold Member
MHB
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Hello! 
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)
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)
