MHB Finding equation normal to a plane and certain point

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To find the equation of a line normal to the plane defined by x + 2y + 3z = 12 and passing through the point (4,6,8), the normal vector is identified as (1,2,3). The line can be expressed using the formula l(t) = a + tb, where 'a' is the point (4,6,8) and 'b' is the normal vector (1,2,3). Substituting these values gives the line equation as l(t) = (4,6,8) + t(1,2,3). The discussion concludes with the original poster successfully deriving the line equation.
brunette15
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For the following question I am given a plane: x + 2y + 3z = 12. I want to find the equation of a line normal to the plane and going through the point (4,6,8). I am trying to use the formula N . (r - r0) = 0 however seem to be getting the incorrect answer :(
 
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The line equation that passes through $a$ and is parallel to $b$ is $l(t)=a+tb$.

From the equation of the plane $x + 2y + 3z = 12$, we have that a normal vector to the plane is $(1,2,3)$.

We are looking for a line that is normal to the plane, so parallel to the vector $(1,2,3)$ and passes through the point $(4,6,8)$.

Can you find now the line equation?
 
evinda said:
The line equation that passes through $a$ and is parallel to $b$ is $l(t)=a+tb$.

From the equation of the plane $x + 2y + 3z = 12$, we have that a normal vector to the plane is $(1,2,3)$.

We are looking for a line that is normal to the plane, so parallel to the vector $(1,2,3)$ and passes through the point $(4,6,8)$.

Can you find now the line equation?

Thankyou so much I was able to figure it out from there :)
 
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