What is the equation of the line passing thro | vedclass

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(B) The line passes through the point $(x_1, y_1, z_1) = (1, 2, 3)$.Since the line is perpendicular to the plane $3x + 4y - 5z = 6$,the direction vect

Vedclass · Accepted Answer

$\frac{x - 1}{3} = \frac{y - 2}{4} = \frac{3 - z}{5}$

Vedclass · Answer

(B) The line passes through the point $(x_1, y_1, z_1) = (1, 2, 3)$.Since the line is perpendicular to the plane $3x + 4y - 5z = 6$,the direction vector of the line is the same as the normal vector of the plane,$\vec{n} = (3, 4, -5)$.The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.Substituting the values,we get $\frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - 3}{-5}$.This can be rewritten as $\frac{x - 1}{3} = \frac{y - 2}{4} = \frac{3 - z}{5}$.

What is the equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $3x + 4y - 5z = 6$?

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