The value of m,such that the line frac{x-4}{1 | vedclass

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(A) The line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-m}{2}$ lies in the plane $2x-4y+z=7$.First,the point $(4, 2, m)$ lies on the line,so it must also li

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$7$

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(A) The line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-m}{2}$ lies in the plane $2x-4y+z=7$.First,the point $(4, 2, m)$ lies on the line,so it must also lie on the plane.Substituting the point $(4, 2, m)$ into the plane equation $2x-4y+z=7$:$2(4) - 4(2) + m = 7$$8 - 8 + m = 7$$m = 7$Second,the direction vector of the line $\vec{v} = (1, 1, 2)$ must be perpendicular to the normal vector of the plane $\vec{n} = (2, -4, 1)$.Checking the dot product: $\vec{v} \cdot \vec{n} = (1)(2) + (1)(-4) + (2)(1) = 2 - 4 + 2 = 0$.Since the dot product is $0$,the line is parallel to the plane. Since the point $(4, 2, 7)$ lies on the plane,the entire line lies in the plane.Thus,the value of $m$ is $7$.

The value of $m$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-m}{2}$ lies in the plane $2x-4y+z=7$,is

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