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The vector equation of the line passing through the point (-1, 5, 4) and perpendicular to the plane z=0 is : |
$\vec{r}= -\hat{i}+5\hat{j} + 4\hat{k} + \lambda (\hat{i}+\hat{j})$ $\vec{r}= -\hat{i}+5\hat{j} + (-4 + \lambda) \hat{k}$ $\vec{r}=-\hat{i}+5\hat{j}+ 4\hat{k} + \lambda \hat{k}$ $\vec{r}= \lambda \hat{k}$ |
$\vec{r}=-\hat{i}+5\hat{j}+ 4\hat{k} + \lambda \hat{k}$ |
The correct answer is Option (3) → $\vec{r}=-\hat{i}+5\hat{j}+ 4\hat{k} + \lambda \hat{k}$ $\vec a=-\hat i+5\hat j+4\hat k$ $\vec n=\hat k$ (to plane $z=0$) So line is parallel to $\vec n$ so $\vec r=\vec a+λ\vec n$ eq: $\vec r=-\hat i+5\hat j+4\hat k+λ\hat k$ |
