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

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(A) For the line to lie in the plane,every point on the line must satisfy the equation of the plane.The line passes through the point $(4, 2, k)$ and

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) For the line to lie in the plane,every point on the line must satisfy the equation of the plane.The line passes through the point $(4, 2, k)$ and has a direction vector $\vec{v} = (1, 1, 2)$.First,the point $(4, 2, k)$ must lie on the plane $2x - 4y + z = 7$:$2(4) - 4(2) + k = 7$$8 - 8 + k = 7$$k = 7$Second,the direction vector of the line $\vec{v} = (1, 1, 2)$ must be perpendicular to the normal vector of the plane $\vec{n} = (2, -4, 1)$:$\vec{v} \cdot \vec{n} = (1)(2) + (1)(-4) + (2)(1) = 2 - 4 + 2 = 0$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 $k$ is $7$.

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

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