0
$\begingroup$

I have:

Plane: $2x+2y-z=1$

Line: $(1,1,0)+t(-1,-1,2)$

How to get the point $p$ which is in the plan and how to know the distance ? i know the normal vector is $(2,2-1)$.

Any help will be appreciated !

$\endgroup$
3
  • 1
    $\begingroup$ Hint: It looks like your line and plane intersect. $\endgroup$
    – mickep
    Mar 10, 2015 at 15:06
  • $\begingroup$ Nope, i dont think so. $\endgroup$
    – hussi
    Mar 10, 2015 at 15:56
  • $\begingroup$ The line and plane you have written intersect. Maybe there is an error. The answer by @thelionkingrafiki is correct. $\endgroup$
    – mickep
    Mar 10, 2015 at 15:57

2 Answers 2

1
$\begingroup$

Since $(2,2,-1) \cdot (-1,-1,2) = -2-2-2=-6 \neq 0$, the normal vector is not perpendicular to the vector of the line, the line must cross the plane. So their distance is $0$.

Now, if you are interested in the intersection point you need to solve:

$$2(1-t) + 2(1-t)-2t = 1$$ which gives $t = \frac 1 2$, and thus the intersection point is $(0.5, 0.5, 1)$.

$\endgroup$
1
  • $\begingroup$ they are not intersect eachother. $\endgroup$
    – hussi
    Mar 10, 2015 at 15:55
0
$\begingroup$

Let's write the equation of the plane in the form $$ \cos\alpha x+ \cos\beta y + \cos\gamma z + p = 0, $$ where ${\cos^2\alpha+\cos^2\beta+\cos^2\gamma}=1$, then for any point $(x^*,y^*,z^*)$ $$|\cos\alpha x^*+ \cos\beta y^* + \cos\gamma z^* + p|$$ will be the distance between the point and the plane.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .