The cartesian equation of a line $2 x-3=3 y+1=5-6 z$. The vector equation of the line passing through the…

The cartesian equation of a line $2 x-3=3 y+1=5-6 z$. The vector equation of the line passing through the point $(7,-5,0)$ and parallel to the given line is
  1. $r=(5 \hat{i}-7 \hat{j})+\lambda(3 \hat{i}+2 \hat{j}-\hat{k})$
  2. $r=(7 \hat{i}+5 \hat{j})+\lambda(3 \hat{i}-2 \hat{j}+\hat{k})$
  3. $r=(7 \hat{i}-5 \hat{j})+\lambda(3 \hat{i}+2 \hat{j}-\hat{k})$
  4. $r=(-5 \hat{i}+7 \hat{j})+\lambda(-3 \hat{i}-2 \hat{j}-\hat{k})$

Solution

Equation of given line is $ \begin{aligned} & 2 x-3=3 y+1=5-6 z \\ & \frac{x-\frac{3}{2}}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{5}{6}}{-1} \end{aligned} $ $\therefore$ Equation of line passes through point $(7,-5,0)$ and having paralleI vector $3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ is $ \mathbf{r}=(7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}})+\lambda(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) $

Asked in: AP EAMCET 2020 (22 Sep Shift 1)

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