Find the vector equation of the line which is parallel to the vector $3\hat{i} - 2\hat{j} + 6\hat{k}$ and which passes through the point $(1, -2, 3)$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} - 2\hat{j} + 6\hat{k})$ ##
Let $\mathbf{a} = 3\hat{i} - 2\hat{j} + 6\hat{k}$ and $\mathbf{b} = \hat{i} - 2\hat{j} + 3\hat{k}$
So, vector equation of the line, which is parallel to the vector $\mathbf{a} = 3\hat{i} - 2\hat{j} + 6\hat{k}$ and passes through the vector $\mathbf{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ is $\mathbf{r} = \mathbf{b} + \lambda \mathbf{a}$.
$∴\mathbf{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} - 2\hat{j} + 6\hat{k})$
$\Rightarrow (x\hat{i} + y\hat{j} + z\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = \lambda(3\hat{i} - 2\hat{j} + 6\hat{k})$
$\Rightarrow (x - 1)\hat{i} + (y + 2)\hat{j} + (z - 3)\hat{k} = \lambda(3\hat{i} - 2\hat{j} + 6\hat{k})$