# Equal intercept on the three axes

• arham_jain_hsr
arham_jain_hsr
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.
$\implies$ line is along the vector $a(\hat i + \hat j + \hat k)$
$\implies$ Equation of line is $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{1}$"

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?

arham_jain_hsr said:
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.
$\implies$ line is along the vector $a(\hat i + \hat j + \hat k)$
$\implies$ Equation of line is $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{1}$"

As per my understanding, an intercept is the value of x, y or z where a line "intersects" the axes.
That's what I would think.
arham_jain_hsr said:
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?
That line given as the solution does not intersect any of the coordinate axes.

arham_jain_hsr said:
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?
I can't see how a line can intersect all three axes, either, except at the origin. It's almost as if they were really asking about a plane with those three intercepts, and then came up with a solution that is a line. Weird...

arham_jain_hsr said:
Homework Statement: Find the equation of a line which passes through the point (2,3,4) and which has equal intercepts on the axes.
The homework statement is badly stated, because it's trying to ask for an equation of a line in ##\mathbb{R}^3## whose projections onto the ##xy##, ##xz##, and ##yz## planes have 'equal intercepts', meaning each plane intersects two axis at equivalent values, up to but not including signs. One mistake in the statement is that 'equal intercepts' is wrong. It should be 'equal absolute values of the intercepts'. If you try to construct an intersection of three planes whose projections onto the ##xy, yz, xz## planes have equal intercepts including signs, you can prove that the three planes always intersect at a point. Also, you can show that a line with 'equal intercepts' that crosses ##(a,b,c)## has the parametric form ##(a,b,c)+(\pm1,\pm1,\pm1)t##, so technically there are ##2^3=8## correct answers. For the record, the line given by the book is an intersection of a plane that intercept the ##x## at ##-2## and the ##z## axis at ##2##, a plane that intercepts the ##y## axis at ##-1## and the ##z## axis at ##1##, and a plane that intercepts the ##x## at ##-1## and the ##y## axis at ##1##. The problem is in 3D, so you can use a 3-D grapher.

Edit: I didn't prove each of these statements, and if you want to to use them, you should prove them first, to be sure.

Last edited:
Well, reverse engineering, we get the solution line :
## x=2+t##
##y=3+t##
##z=4+t##
So that it's the line through ##(2,3,4)## in the direction of ##(1,1,1)##. Not sure how, what this agrees with " having equal intercept with the axes".

docnet
WWGD said:
Well, reverse engineering, we get the solution line :
## x=2+t##
##y=3+t##
##z=4+t##
So that it's the line through ##(2,3,4)## in the direction of ##(1,1,1)##. Not sure how, what this agrees with " having equal intercept with the axes".
It doesn't agree, you're right. But sometimes the poor students have to 'connect the dots' to make sense of a problem if a professor's (or a textbook's) problem doesn't make any sense as stated. It happens rarely but this case seems to be one.

Source: I'm a poor student.

WWGD
docnet said:
It doesn't agree, you're right. But sometimes the poor students have to 'connect the dots' to make sense of a problem if a professor's (or a textbook's) problem doesn't make any sense as stated. It happens rarely but this case seems to be one.

Source: I'm a poor student.
And , if I did the computations correctly, the angles between the line and the xz, xy, yz axes are not all equal either.

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