The vector equation of the line 2x+4=3y+1=6z- | vedclass

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(B) The given equation of the line is $2x+4=3y+1=6z-3$. To convert this into the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$,we re

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

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(B) The given equation of the line is $2x+4=3y+1=6z-3$. To convert this into the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$,we rewrite the equation as: $2(x+2)=3(y+\frac{1}{3})=6(z-\frac{1}{2})$. Dividing throughout by the least common multiple of $2, 3, 6$,which is $6$,we get: $\frac{2(x+2)}{6}=\frac{3(y+\frac{1}{3})}{6}=\frac{6(z-\frac{1}{2})}{6} \Rightarrow \frac{x+2}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{2}}{1}$. Thus,the line passes through the point $(-2, -\frac{1}{3}, \frac{1}{2})$ and has direction ratios $(3, 2, 1)$. The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$. Therefore,the vector equation is $\overline{r}=\left(-2 \hat{i}-\frac{1}{3} \hat{j}+\frac{1}{2} \hat{k}\right)+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})$.

The vector equation of the line $2x+4=3y+1=6z-3$ is

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