The discussion centers on finding the equation of a line normal to the plane defined by the equation x + 2y + 3z = 12, which passes through the point (4, 6, 8). The normal vector to the plane is identified as (1, 2, 3). The correct line equation is derived using the formula l(t) = a + tb, where 'a' is the point (4, 6, 8) and 'b' is the normal vector (1, 2, 3). The final line equation is l(t) = (4, 6, 8) + t(1, 2, 3).
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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?