Find the vector equation of the line passing through the point $(2, 3, -5)$ and making equal angles with the coordinate axes.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \pm \frac{k}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$ ##
Given $\alpha = \beta = \gamma$
Now, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
$3\cos^2\alpha = 1 \text{ or, } 3\cos^2\beta = 1$
$3\cos^2\alpha = 1 \text{ or, } \cos^2\alpha = \frac{1}{3} \quad [∵\alpha = \beta = \gamma]$
or, $\cos\alpha = \pm \frac{1}{\sqrt{3}}$
Thus, $\cos\alpha = \cos\beta = \cos\gamma = \pm \frac{1}{\sqrt{3}}$
$= l = m = n$
Let the required equation of line is
$\vec{r} = \vec{a} + k\vec{b}$
Here,
$\vec{a} = 2\hat{i} + 3\hat{j} - 5\hat{k}$
$\vec{b} = \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$
Thus, the required equation is
$\vec{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \pm \frac{k}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$