Appeared in: 2024 (6 sets)
Find the vector equation of the line passing through the point $(2, 3, -5)$ and making equal angles with the co-ordinate axes.
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Official Evaluation Steps
Step 1 (Formula / Governing Law):
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:
$$\vec{r} = \vec{a} + \lambda \vec{b}$$Where $\vec{a} = 2\hat{i} + 3\hat{j} - 5\hat{k}$. Since the line makes equal angles ($\alpha$) with the coordinate axes, the direction cosines $l, m, n$ are equal ($l = m = n = \cos \alpha$). (0.5 Mark)
Step 2 (Step-by-Step Working):
Using the property of direction cosines, $l^2 + m^2 + n^2 = 1$:
$$l^2 + l^2 + l^2 = 1 \implies 3l^2 = 1 \implies l = \pm \frac{1}{\sqrt{3}}$$Thus, the direction ratios of the line are proportional to $(1, 1, 1)$. Therefore, the vector $\vec{b}$ parallel to the line is $\hat{i} + \hat{j} + \hat{k}$. (1.0 Mark)
Step 3 (Final Answer):
Substituting $\vec{a}$ and $\vec{b}$ into the vector equation formula:
$$\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$$Where $\lambda$ is a scalar. (0.5 Mark)