Question

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

• A. $\vec{r} = \left(3\hat{i}+7\hat{j}+2\hat{k}\right)+\mu \left(5\hat{i}+4j-6\hat{k}\right)$
• B. $\vec{r} = \left(5\hat{i}+4\hat{j}-6\hat{k}\right)+\mu \left(3\hat{i}+7j+2\hat{k}\right)$
• C. $\vec{r} = \left(5\hat{i}-4\hat{j}-6\hat{k}\right)+\mu \left(3\hat{i}-7j-2\hat{k}\right)$
• D. $\vec{r} = \left(5\hat{i}-4\hat{j}+6\hat{k}\right)+\mu \left(3\hat{i}+7j+2\hat{k}\right)$

Answer 1

Answer: (D)

$ \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$ have vector form
$= \left(x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}\right)+\lambda\left(a\hat{i}+b\hat{j}+c\hat{k}\right)$
Required equation in vector form is
$ \vec{r} = \left(5\hat{i}-4\hat{j}+6\hat{k}\right)+\mu \left(3\hat{i}+7j+2\hat{k}\right)$