The point where the line frac{x - 1}{2} = fra | vedclass

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(A) Let the general point on the line be $(x, y, z) = (2k + 1, -3k + 2, 4k - 3)$,where $k$ is a constant.Since this point lies on the plane $2x + 4y -

Vedclass · Accepted Answer

$(3, -1, 1)$

Vedclass · Answer

(A) Let the general point on the line be $(x, y, z) = (2k + 1, -3k + 2, 4k - 3)$,where $k$ is a constant.Since this point lies on the plane $2x + 4y - z = 1$,we substitute the coordinates into the plane equation:$2(2k + 1) + 4(-3k + 2) - (4k - 3) = 1$$4k + 2 - 12k + 8 - 4k + 3 = 1$$-12k + 13 = 1$$-12k = -12$$k = 1$Substituting $k = 1$ back into the point coordinates:$x = 2(1) + 1 = 3$$y = -3(1) + 2 = -1$$z = 4(1) - 3 = 1$Thus,the required point is $(3, -1, 1)$.

The point where the line $\frac{x - 1}{2} = \frac{y - 2}{-3} = \frac{z + 3}{4}$ meets the plane $2x + 4y - z = 1$ is:

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