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

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(B) For a line to lie in a plane,every point on the line must satisfy the equation of the plane. The line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z+m}{2}$

Vedclass · Accepted Answer

$-7$

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(B) For a line to lie in a plane,every point on the line must satisfy the equation of the plane. The line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z+m}{2}$ passes through the point $(4, 2, -m)$.Substituting this point into the plane equation $2x - 4y + z = 7$:$2(4) - 4(2) + (-m) = 7$$8 - 8 - m = 7$$-m = 7$$m = -7$Additionally,the normal vector of the plane $\vec{n} = (2, -4, 1)$ must be perpendicular to the direction vector of the line $\vec{v} = (1, 1, 2)$.Checking the dot product: $(2)(1) + (-4)(1) + (1)(2) = 2 - 4 + 2 = 0$.Since the dot product is $0$,the line is parallel to the plane. Since the point $(4, 2, -m)$ lies on the plane for $m = -7$,the entire line lies in the plane.

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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