The Cartesian equation of the line which is p | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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