The equation of a line passing through the po | vedclass

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(B) The given equation of the line is $3x + 4 = 4y - 1 = 1 - 4z$. We rewrite this in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_

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$\frac{x-2}{4} = \frac{y-4}{3} = \frac{z-6}{-3}$

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(B) The given equation of the line is $3x + 4 = 4y - 1 = 1 - 4z$. We rewrite this in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ by dividing by the coefficients of $x, y, z$: $\frac{x + 4/3}{1/3} = \frac{y - 1/4}{1/4} = \frac{z - 1/4}{-1/4}$. Multiplying the denominators by $12$,we get the direction ratios $(a, b, c)$ as $(4, 3, -3)$. The equation of a line passing through $(x_0, y_0, z_0) = (2, 4, 6)$ with direction ratios $(4, 3, -3)$ is: $\frac{x-2}{4} = \frac{y-4}{3} = \frac{z-6}{-3}$.

The equation of a line passing through the point $(2, 4, 6)$ and parallel to the line $3x + 4 = 4y - 1 = 1 - 4z$ is

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