Efficient solution to a 3D geometry distance problem

[Krushnaraj Pandya]

Homework Statement

Find the coordinates of those points on the lines (x+1)/2 = (y+2)/2 = (z-3)/6 which is at a distance of 3 units from the point (1,-2,3)

2. Relevant methods
1) assume a point, use distance formula- (very calculative)
2) write vector equation of line, find foot of perpendicular (k) from point (t) to line, assume the point (m)- use cross product to write sinθ in terms of these points- (again, very calculative)

The Attempt at a Solution

So I'm looking for a solution that one might actually be able to employ to solve this in a time-bound competitive exam with less hassle and tediousness. Would be glad to know new ideas; thank you.

 

[Mark44]

Homework Statement

Find the coordinates of those points on the lines (x+1)/2 = (y+2)/2 = (z-3)/6 which is at a distance of 3 units from the point (1,-2,3)

2. Relevant methods
1) assume a point, use distance formula- (very calculative)
2) write vector equation of line, find foot of perpendicular (k) from point (t) to line, assume the point (m)- use cross product to write sinθ in terms of these points- (again, very calculative)

The Attempt at a Solution

So I'm looking for a solution that one might actually be able to employ to solve this in a time-bound competitive exam with less hassle and tediousness. Would be glad to know new ideas; thank you.

As I said before, all of the techniques require a lot of calculations, so I don't think there is any "magic bullet," other than to work a lot of problems beforehand. Test yourself on problems where the answers are provided. I think it's better to finish 8 problems out of 10 that are done correctly than to work all 10 problems, but with errors on half of therm.

 

[Ray Vickson]

Homework Statement

Find the coordinates of those points on the lines (x+1)/2 = (y+2)/2 = (z-3)/6 which is at a distance of 3 units from the point (1,-2,3)

2. Relevant methods
1) assume a point, use distance formula- (very calculative)
2) write vector equation of line, find foot of perpendicular (k) from point (t) to line, assume the point (m)- use cross product to write sinθ in terms of these points- (again, very calculative)

The Attempt at a Solution

So I'm looking for a solution that one might actually be able to employ to solve this in a time-bound competitive exam with less hassle and tediousness. Would be glad to know new ideas; thank you.

You can fairly quickly get a quadratic equation to solve. If you set the coordinates on the line as $(x,y,z) = (-1 + 2t, -2+2 t, 3 + 6t)$, you want to find the value of $t$ that has $(x,y,z)$ at a distance of 3 from the point $(1,-2,3)$, or whose squared distance is 9 units. The squared distance is $S = [1-(-1+2t)]^2 + [-2 -(-2+2t)]^2 + [3 - (3 + 6t)]^2\\ \hspace{2ex}= (2 - 2t)^2 + (-2t)^2 +(-6t)^2$ You an easily expand out $S$ to get a quadratic of the form $S = a t^2 + bt + c$, then solve your quadratic equation $a t^2 + bt +c = 9.$

As a bonus, available with almost no extra work: if the two roots of the quadratic are $t_1$ and $t_2$, the mid-point $t_m = \frac{1}{2}(t_1+t_2)$ gives the point on the line that is closest to the given point $(1,-2,3).$ Note that we can also get this point without knowing $t_1$ and $t_2$ because the sum of the roots is closely related to the quadratic coefficients.

 

[ehild]

You can fairly quickly get a quadratic equation to solve. If you set the coordinates on the line as $(x,y,z) = (-1 + 2t, -2+2 t, 3 + 6t)$, you want to find the value of $t$ that has $(x,y,z)$ at a distance of 3 from the point $(1,-2,3)$, or whose squared distance is 9 units.

Those points are the intersections of the lines with the circle of center (1, -2, 3) and radius 3.

 

[Krushnaraj Pandya]

Thank you everyone for your help, and sorry that I'm taking so long to get back- I'm in a mid-exam rush and will respond after understanding everyone's replies properly. Thank you again :D

 

[Krushnaraj Pandya]

As a bonus, available with almost no extra work: if the two roots of the quadratic are t1t1t_1 and t2t2t_2, the mid-point tm=12(t1+t2)tm=12(t1+t2)t_m = \frac{1}{2}(t_1+t_2) gives the point on the line that is closest to the given point (1,−2,3).(1,−2,3).(1,-2,3).

I can't visualize why this happens, but it is a very beautiful solution. Thank you

Those points are the intersections of the lines with the circle of center (1, -2, 3) and radius 3.

Seems logical since the locus of the point at 3 units is a circle and the line is simply a tangent.

As I said before, all of the techniques require a lot of calculations, so I don't think there is any "magic bullet," other than to work a lot of problems beforehand. Test yourself on problems where the answers are provided. I think it's better to finish 8 problems out of 10 that are done correctly than to work all 10 problems, but with errors on half of therm.

True, Although withput @Ray Vickson's solution I would have taken a huge amount of time to get to the right place. This is by far the easiest and most productive solutions and the kind that can really help me. Of course I'll practice though...I'm already on the slower side in exams which is a huge drawback

 

[ehild]

Seems logical since the locus of the point at 3 units is a circle and the line is simply a tangent.

In general, the line is a secant.