Question

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

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

Answer 1

Answer: (D)

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$, having vector form
$\vec{r}=\left(x_1\hat{i}+y_1\hat{j}+z_1\hat{k}\right)+\lambda \left(a\hat{i}+b\hat{j}+c\hat{k}\right)$
$\therefore$ Required equation in vector form is
$\vec{r}=\left(5\hat{i}-4\hat{j}+6\hat{k}\right)+\mu \left(3\hat{i}+7\hat{j}+2\hat{k}\right)$