- #1

arham_jain_hsr

- 23

- 7

- Homework Statement
- Find the equation of a line which passes through the point (2,3,4) and which has equal intercepts on the axes.

- Relevant Equations
- 3-dimensional coordinate geometry

The solution in my book is as follows:

"Since the line has equal intercepts on axes, it is equally inclined to axes.

[itex]\implies[/itex] line is along the vector [itex]a(\hat i + \hat j + \hat k)[/itex]

[itex]\implies[/itex] Equation of line is [itex]\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{1}[/itex]"

As per my understanding, an intercept is the value of x, y or z where a line "intersects" the axes. How come in the solution does having equal "intercept" on axes imply an equal "inclination" to the axes? If my understanding is correct, I can't seem to think of any possible way a line can have an equal intercept on "all" three axes other than at 0. Am I missing something out?

"Since the line has equal intercepts on axes, it is equally inclined to axes.

[itex]\implies[/itex] line is along the vector [itex]a(\hat i + \hat j + \hat k)[/itex]

[itex]\implies[/itex] Equation of line is [itex]\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{1}[/itex]"

As per my understanding, an intercept is the value of x, y or z where a line "intersects" the axes. How come in the solution does having equal "intercept" on axes imply an equal "inclination" to the axes? If my understanding is correct, I can't seem to think of any possible way a line can have an equal intercept on "all" three axes other than at 0. Am I missing something out?