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Question

(A) The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and parallel to a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is

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$\frac{x-5}{3} = \frac{y+2}{-2} = \frac{z-4}{8}$

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(A) The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and parallel to a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by the formula: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Given the point $(x_1, y_1, z_1) = (5, -2, 4)$ and the direction vector components $(a, b, c) = (3, -2, 8)$.Substituting these values into the formula, we get: $\frac{x-5}{3} = \frac{y-(-2)}{-2} = \frac{z-4}{8}$.Simplifying the expression, we obtain: $\frac{x-5}{3} = \frac{y+2}{-2} = \frac{z-4}{8}$.

The Cartesian equation of the line passing through the point $(5, -2, 4)$ and parallel to the vector $3\hat{i}-2\hat{j}+8\hat{k}$ is . . . . . . .

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