If the Cartesian equation of a line is 6x-2=3 | vedclass

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(A) The given Cartesian equation of the line is $6x-2=3y+1=2z-2$. We rewrite this in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c

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

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(A) The given Cartesian equation of the line is $6x-2=3y+1=2z-2$. We rewrite this in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ by factoring out the coefficients of $x, y, z$: $6(x-\frac{1}{3})=3(y+\frac{1}{3})=2(z-1)$. Dividing the entire equation by the least common multiple of $6, 3, 2$,which is $6$,we get: $\frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-1}{3}$. This line passes through the point $A(\frac{1}{3}, -\frac{1}{3}, 1)$ and has direction ratios proportional to $\vec{v} = \hat{i}+2\hat{j}+3\hat{k}$. The vector equation of a line passing through point $\vec{a}$ with direction $\vec{v}$ is $\vec{r} = \vec{a} + \lambda \vec{v}$. Substituting the values,we get $\vec{r} = (\frac{1}{3}\hat{i} - \frac{1}{3}\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$.

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

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