I have been working on this exercise and am kinda struggling with it. This is the exercise and what I have done so far. Any tips would be greatly appreciated!

Given the plane $x+y=0$ and two lines: $p_1: \frac{x}{3} = \frac{y+1}{1} = \frac{z-3}{-2}$ and $p_2$ (form with 2 planes): $ y=z+2$ & $x=1$ find the line $q$ that is parallel to the first plane and it intersects $p_1$ and $p_2$ in two points which distance is 3.

This is what I have so far. First, I transformed $p_2$ to canonical form: cross product of two normal vectors of given planes will give a direction vector for $p_2$ . $\vec{n_1} =(0,1,-1)$ and $\vec{n_2}=(1,0,0)$

Now for the cross product: $\vec{n_1} \times \vec{n_2} = (0,-1,-1).$

I can now choose a point that satisfies both planes of $p_2$. For e.g. $A=(1,3,1)$. By the formula, now I have a canonical line form: $p_2:\frac{x-1}{0}=\frac{y-3}{-1}=\frac{z-1}{-1}$

I also have the info about distance. Let $T_1=(x_1,y_1,z_1)$ and $T_2=(x_2,y_2,z_2)$. Their distance is 3 so I have: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}=3$. After the transformation I get: $(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2=3$.

Now I already have an equation with 6 unknowns. What can I do now to find the line $q$ ? I can also find form of $q$ because I know that normal vector to the $x+y=0$ is also normal vector to the $q$. Should I write $q$ in a form of $k,l,m$ or in a form of two unknown points? I also can write two determinants of $q$ and $p_1$,$p_2$ and set them equal to 0 but I don't get enough info.

Can someone please give me a hint or help? Sorry if formatting is not good, I'm trying my best. Have a nice day!


3 Answers 3



Given the plane $\Pi$ and the lines $L_1,\ L_2$ as

$$ \cases{ \Pi\to (p-p_0)\cdot \vec n = 0\\ L_1\to p = p_1 +\lambda_1\vec v_1\\ L_2\to p = p_2 +\lambda_2\vec v_2 } $$

the generic plane parallel to $\Pi$ can be defined as

$$ \Pi_p\to (p-p_0)\cdot \vec n = c\\ $$

Now if $\Pi_c\cap L_1\ne 0$ and $\Pi_c\cap L_2\ne 0$ we have

$$ \cases{ (p_1-p_0+\lambda_1\vec v_1)\cdot \vec n = c\\ (p_2-p_0+\lambda_2\vec v_1)\cdot \vec n = c\\ ||p_1-p_2+\lambda_1\vec v_1-\lambda_2\vec v_2|| = d } $$

Here we have three equations and three unknowns $\lambda_1,\lambda_2,c$


Say the point where it intersects $p_1$ is $$P_1=\left(3a,a-1,-2a+3\right)$$and where it intersects $p_2$ is $$P_2=\left(1,-b+3,-b+1\right)$$Then we know the vector $\vec{P_1P_2}$ is parallel to $q$, which is further perpendicular to the normal vector of $x+y=0$. So, $$(3a-1, a+b-4,-2a+b+2)\cdot(1,1,0) = 0 \\ 3a-1 +a+b-4=0 \\ \implies 4a+b=5$$ Also, $|P_1P_2| =3$, i.e. $$\sqrt{(3a-1)^2 +(a+b-4)^2 +(-2a+b+2)^2} =3$$ Solve the two equations to get $a,b$ and then you have two points through which $q$ passes, and you can write its equation.

  • $\begingroup$ that helped me a lot, I have solved it! cheers $\endgroup$ Jul 13, 2020 at 20:43
  • 1
    $\begingroup$ @Exzone Awesome. $\endgroup$
    – Vishu
    Jul 13, 2020 at 21:16
  • $\begingroup$ Just wanted to say thanks, I've just passed my Analytic geometry exam. We were given the same type of exercise! $\endgroup$ Jul 16, 2020 at 10:15
  • $\begingroup$ @Exzone Always glad to help! $\endgroup$
    – Vishu
    Jul 16, 2020 at 10:16

It's a vector problem of a kind "Express (all given conditions algebraically) and solve (a system of equations), don't think."
First, we'll need the second line equation in the parametric form.
Having $\ell_1=\mathbf{a}_1+t\mathbf{d}_1$ and $\ell_2=\mathbf{a}_2+s\mathbf{d}_2$ we will have $2$ equations: $|\ell_1(t)-\ell_2(s)|=3$ and $(\ell_1(t)-\ell_2(s))\cdot\mathbf{n}=0$ where $\mathbf{n}$ is the normal vector of the given plane.
By finding $s,\,t$ we will be able to construct a line through points $\mathbf{a}=\ell_1(t)$, $\mathbf{b}=\ell_2(s)$ in the form $\mathbf{x}= \mathbf{a}+u(\mathbf{b}-\mathbf{a})$.

Let's proceed:
In general case, going from "two planes intersection" equation of a line to paranetric form involves cross product of planes' normal vectors. Here we don't need it as we can simply write $y=z+2=s,\,x=1$ so $\ell_2(s)=(1,0,-2)+s(0,1,1)$.
The first line parametric form equation can be obtained by $$\frac{x}{3}=\frac{y+1}{1}=\frac{z-3}{-2}=t$$ $$\begin{cases} x=3t\\y=-1+t\\z=3-2t \end{cases}$$ $$\ell_1(t)=(0,-1,3)+t(3,1,-2)$$ Now $\ell_2(s)-\ell_1(t)=(1,1,-5)+s(0,1,1)+t(-3,-1,2)$ and $9=(\ell_2(s)-\ell_1(t))^2$ becomes $$2 s^2 + 2 s t - 8 s + 14 t^2 - 28 t + 27=9\tag{1}$$ and (because for a plane $Ax+By+Cz-D=0$ the normal vector is $(A,B,C)$, here ($x+y=0$) the normal vector is $(1,1,0)$) $$((1,1,-5)+s(0,1,1)+t(-3,-1,2))\cdot (1,1,0)=0$$ $$s - 4 t + 2=0$$ hence substituting $s=4t-2$ into (1) and we have a quadratic with respect to $t$, $$27 t^2 - 48 t + 21 = 0$$ solving which we have $$t_1 = \frac{7}{9},\ t_2=1$$ and thus $$s_1=\frac{10}{9},\ s_2=2$$ For the first case the points are $$\left(\frac{7}{3}, -\frac{2}{9}, \frac{13}{9}\right), \left(1, \frac{10}{9}, -\frac{8}{9}\right)$$ and the line is $$\frac{x-\frac{7}{3}}{-4}=\frac{y+\frac{2}{9}}{4}=\frac{z-\frac{13}{9}}{-7},$$ for the second case the points are $$(3, 0, 1),\, (1, 2, 0)$$ and the line equation is $$\frac{x-3}{-2}=\frac{y}{2}=\frac{z-1}{-1}.$$ (Numerical computations here).

  • $\begingroup$ By doing what Mr. Tavish said below I got a line. Thanks for the effort and help, I appreciate it a lot! $\endgroup$ Jul 13, 2020 at 20:44
  • $\begingroup$ Did edit the answer the third time) Can you please write down your resulting line equations? (even if they differ from above) so I can see if mine are correct. Thanks. $\endgroup$ Jul 13, 2020 at 21:01
  • $\begingroup$ Yes. I got the exact same result. Thanks so much!! $\endgroup$ Jul 13, 2020 at 21:05

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