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If cartesian equation of the line is $x-1=2 y+3=3-z$, then its vector equation
is
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is
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Verified Answer
The correct answer is:
$\bar{r}=\left(\hat{\imath}-\frac{3}{2} \hat{\jmath}+3 \hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})$
Given Cartesian equation is
$\begin{aligned}
x-1 &=2 y+3=3-z \\
\therefore \frac{x-1}{1} &=\frac{2\left(y+\frac{3}{2}\right)}{1}=\frac{-(z-3)}{1} \Rightarrow \frac{x-1}{1}=\frac{y+\frac{3}{2}}{\left(\frac{1}{2}\right)}=\frac{z-3}{-1}
\end{aligned}$
$\begin{array}{l}
\therefore 1, \frac{1}{2},-1 \quad \text { i.e. } 2,1,-2 \text { are d.r. of given line. } \\
\therefore \quad \mathrm{A}\left(1,-\frac{3}{2}, 3\right) \text { lies on given line. } \\
\therefore \quad \text { vector equation is } \\
\qquad \overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})
\end{array}$
$\begin{aligned}
x-1 &=2 y+3=3-z \\
\therefore \frac{x-1}{1} &=\frac{2\left(y+\frac{3}{2}\right)}{1}=\frac{-(z-3)}{1} \Rightarrow \frac{x-1}{1}=\frac{y+\frac{3}{2}}{\left(\frac{1}{2}\right)}=\frac{z-3}{-1}
\end{aligned}$
$\begin{array}{l}
\therefore 1, \frac{1}{2},-1 \quad \text { i.e. } 2,1,-2 \text { are d.r. of given line. } \\
\therefore \quad \mathrm{A}\left(1,-\frac{3}{2}, 3\right) \text { lies on given line. } \\
\therefore \quad \text { vector equation is } \\
\qquad \overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})
\end{array}$
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