Find the equation of a line, that lies on a plane $2 x - 3 y + 4 z = 8$ $2x-3y+4z=8$ , perpendicular to a line $\frac{x - 1}{3} = \frac{y - 1}{2} = \frac{z - 2}{3}$ $(x-1)/3=(y-1)/2=(z-2)/3$ and passes through the point $(1, 2, - 1)$ $(1,2,-1)$ ?

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Find the equation of a line, that lies on a plane $2 x - 3 y + 4 z = 8$ $2x-3y+4z=8$ , perpendicular to a line $\frac{x - 1}{3} = \frac{y - 1}{2} = \frac{z - 2}{3}$ $(x-1)/3=(y-1)/2=(z-2)/3$ and passes through the point $(1, 2, - 1)$ $(1,2,-1)$ ?

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Answer 1 · 2018-06-23T17:21:58

No such line can exist because the point $(1, 2, - 1)$ $(1,2,-1)$ does not lie on the plane $2 x - 3 y + 4 z = 8$ $2x-3y+4z=8$ , therefore, the line cannot lie on the plane.

Explanation:

An additional reason why no such line can exist is that the direction of the line is $\to l = 3 ˆ i + 2 ˆ j + 3 ˆ k$ $vecl = 3hati+2hatj+3hatk$ ; this means that the only planes that can have lines that intersect the given line perpendicularly must be of the form $3 x + 2 y + 3 z = c$ $3x+2y+3z = c$ where $c$ $c$ is any real number. All other planes cannot have lines that intersect the given line perpendicularly.

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Find the equation of a line, that lies on a plane $2 x - 3 y + 4 z = 8$ $2x-3y+4z=8$ , perpendicular to a line $\frac{x - 1}{3} = \frac{y - 1}{2} = \frac{z - 2}{3}$ $(x-1)/3=(y-1)/2=(z-2)/3$ and passes through the point $(1, 2, - 1)$ $(1,2,-1)$ ?

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Find the equation of a line, that lies on a plane 2x−3y+4z=82x-3y+4z=8, perpendicular to a line x−13=y−12=z−23(x-1)/3=(y-1)/2=(z-2)/3 and passes through the point (1,2,−1)(1,2,-1)?

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Explanation:

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Find the equation of a line, that lies on a plane $2 x - 3 y + 4 z = 8$ $2x-3y+4z=8$ , perpendicular to a line $\frac{x - 1}{3} = \frac{y - 1}{2} = \frac{z - 2}{3}$ $(x-1)/3=(y-1)/2=(z-2)/3$ and passes through the point $(1, 2, - 1)$ $(1,2,-1)$ ?