Practicing Success
The vector equation of a line passing through a point with position vector $2\hat{i}-\hat{j}+\hat{k}$ and parallel to the line joining the points with position vectors $-\hat{i}+4\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+2\hat{k}$ is : |
$\vec{r}=2\hat{i}-\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}+\hat{k})$ $\vec{r}=2\hat{i}-\hat{j}+\hat{k}+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ $\vec{r}=\hat{i}-\hat{j}+\hat{k}+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ $\vec{r}=\hat{i}-2\hat{j}+\hat{k}+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ |
$\vec{r}=2\hat{i}-\hat{j}+\hat{k}+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ |
The correct answer is Option (2) → $\vec{r}=2\hat{i}-\hat{j}+\hat{k}+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ line || vector so vector = $(\hat i+2\hat j+2\hat k)-(-\hat i+4\hat j+\hat k)$ $=2\hat i-2\hat j+\hat k$ Point through which it passes $= 2\hat i-\hat j+\hat k$ so eq. of line = $2\hat{i}-\hat{j}+\hat{k}+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ |