The vector equation of the line frac{x+3}{2}= | vedclass

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(B) The given equation of the line is $\frac{x+3}{2}=\frac{2y-3}{5}; z=-1$. First,rewrite the equation in the standard form $\frac{x-x_1}{a}=\frac{y-y

Vedclass · Accepted Answer

$\bar{r}=\left(-3 \hat{i}+\frac{3}{2} \hat{j}-\hat{k}\right)+\lambda(4 \hat{i}+5 \hat{j})$

Vedclass · Answer

(B) The given equation of the line is $\frac{x+3}{2}=\frac{2y-3}{5}; z=-1$. First,rewrite the equation in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$. $\frac{x+3}{2}=\frac{2(y-3/2)}{5}; z=-1$ $\frac{x-(-3)}{2}=\frac{y-3/2}{5/2}; z=-1$. This line passes through the point $(-3, 3/2, -1)$ and has direction ratios proportional to $(2, 5/2, 0)$. Multiplying the direction ratios by $2$,we get $(4, 5, 0)$. The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$. Here,$\vec{a}=-3\hat{i}+\frac{3}{2}\hat{j}-\hat{k}$ and $\vec{b}=4\hat{i}+5\hat{j}$. Thus,the vector equation is $\vec{r}=\left(-3\hat{i}+\frac{3}{2}\hat{j}-\hat{k}\right)+\lambda(4\hat{i}+5\hat{j})$.

The vector equation of the line $\frac{x+3}{2}=\frac{2y-3}{5}; z=-1$ is

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