Mathematics | VITEEE 2018 | Discussion Page

The vector equation of the symmetrical form of equation of straight line $\frac{x-5}{3} =\frac{y+4}{7} = \frac{z-6}{2}$ is

$\vec{r}=(3\hat{i}+7\hat{j}+2\hat{k})+\mu(5\hat{i}+4j-6\hat{k})$

$\vec{r}=(5\hat{i}+4\hat{j}-6\hat{k})+\mu(3\hat{i}+7j+2\hat{k})$

$\vec{r}=(5\hat{i}-4\hat{j}-6\hat{k})+\mu(3\hat{i}-7j-2\hat{k})$

$\vec{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\mu(3\hat{i}+7j+2\hat{k})$

Option D

$\frac{x-x_{1}}{a} =\frac{y-y_{1}}{b} = \frac{z-z_{1}}{c}$ have vector form

= $(x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k})+\lambda(a\hat{i}+b\hat{j}+c\hat{k})$

Required equation in vector form is

$\vec{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\mu(3\hat{i}+7j+2\hat{k})$

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