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(D) Given the Cartesian equation: $3x + 1 = 6y - 2 = -z + 1$. First,rewrite the equation in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} =

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$\overline{r} = (-\frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} + \hat{k}) + \lambda(2 \hat{i} + \hat{j} - 6 \hat{k})$

Vedclass · Answer

(D) Given the Cartesian equation: $3x + 1 = 6y - 2 = -z + 1$. First,rewrite the equation in the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. $3(x + \frac{1}{3}) = 6(y - \frac{1}{3}) = -(z - 1)$. Dividing by the least common multiple of the coefficients $(6)$,we get: $\frac{x + 1/3}{1/3} = \frac{y - 1/3}{1/6} = \frac{z - 1}{-1}$. To simplify the direction ratios,multiply the denominators by $6$: $\frac{x + 1/3}{2} = \frac{y - 1/3}{1} = \frac{z - 1}{-6}$. The point on the line is $(-\frac{1}{3}, \frac{1}{3}, 1)$ and the direction vector is $2 \hat{i} + \hat{j} - 6 \hat{k}$. Thus,the vector equation is $\overline{r} = (-\frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} + \hat{k}) + \lambda(2 \hat{i} + \hat{j} - 6 \hat{k})$.

The Cartesian equation of a line is $3x + 1 = 6y - 2 = -z + 1$. Find its vector equation.

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