Question

If the Cartesian equation of a line is $6x-2=3y+1=2z-2$, then the vector equation of the line is

• A. $\overline{r}=(\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k})+\lambda(\hat{i}+2\hat{j}+3\hat{k})$
• B. $\overline{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(\hat{i}+2\hat{j}+3\hat{k})$
• C. $\overline{r}=(\frac{-1}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})$
• D. $\overline{r}=(\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$

Hint:

Logic Tip: You can find the direction ratios mentally by covering one part of the equality at a time or taking the reciprocals of the variable coefficients. The coefficients are 6, 3, and 2. Their reciprocals are $1/6, 1/3, 1/2$. Multiply by 6 to clear fractions, yielding direction ratios of $1, 2, 3$.

Answer 1

The Correct Option is A

Concept:
The standard Cartesian form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where $(x_1, y_1, z_1)$ is a point on the line and $(a, b, c)$ are the direction ratios. To convert an equation into this form, the coefficients of $x$, $y$, and $z$ in the numerators must be exactly 1.
Step 1: Factor out the coefficients of the variables.
Given equation: $$6x - 2 = 3y + 1 = 2z - 2$$ Factor out 6, 3, and 2 respectively from each term: $$6\left(x - \frac{2}{6}\right) = 3\left(y + \frac{1}{3}\right) = 2(z - 1)$$ $$6\left(x - \frac{1}{3}\right) = 3\left(y + \frac{1}{3}\right) = 2(z - 1)$$
Step 2: Convert to standard symmetric form.
To remove the coefficients in the numerator, divide the entire equation by the Least Common Multiple (LCM) of 6, 3, and 2, which is 6: $$\frac{6(x - 1/3)}{6} = \frac{3(y + 1/3)}{6} = \frac{2(z - 1)}{6}$$ $$\frac{x - 1/3}{1} = \frac{y + 1/3}{2} = \frac{z - 1}{3}$$
Step 3: Extract the point and direction vector to form the vector equation.
From the standard form, the line passes through the point $(x_1, y_1, z_1) = \left(\frac{1}{3}, -\frac{1}{3}, 1\right)$. The position vector of this point is $\overline{a} = \frac{1}{3}\hat{i} - \frac{1}{3}\hat{j} + \hat{k}$. The direction ratios are $(a, b, c) = (1, 2, 3)$. The direction vector is $\overline{b} = \hat{i} + 2\hat{j} + 3\hat{k}$. The vector equation of the line is $\overline{r} = \overline{a} + \lambda\overline{b}$: $$\overline{r} = \left(\frac{1}{3}\hat{i} - \frac{1}{3}\hat{j} + \hat{k}\right) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$$