If the Cartesian equation of the line is x-1= | vedclass

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(D) The given Cartesian equation is $x-1 = 2y+3 = 3-z$. Rewrite the equation in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}

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

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(D) The given Cartesian equation is $x-1 = 2y+3 = 3-z$. Rewrite the equation in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$: $\frac{x-1}{1} = \frac{y - (-3/2)}{1/2} = \frac{z-3}{-1}$. Thus,the point on the line is $A(1, -3/2, 3)$ and the direction ratios are $(1, 1/2, -1)$. Multiplying the direction ratios by $2$,we get the proportional direction ratios $(2, 1, -2)$. The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r} = \vec{a} + \lambda \vec{b}$. Substituting the values,we get $\vec{r} = (\hat{i} - \frac{3}{2}\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + \hat{j} - 2\hat{k})$.

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

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