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