# 2 lines 1 plane question in 3 dimension

The question I need help with is line $$D: x = {y-2\over -1} = z$$
line $$D': {x-2\over2} = {y-3\over1} = {z+5\over-1}$$
Find plane $$(α)$$ for $$(α)$$ containing $$D$$ and the angle between $$(α)$$ and $$D'$$ is 60 degrees.

This is what I did let $$v,v'$$ be the vector for line $$D,D'$$ respectively. I got $$v = ( 1, -1, 1)$$ ; $$v' = ( 2, 1, -1)$$, which is perpendicular (got from their dot product) meaning $$D$$ and $$D'$$ is perpendicular.

I calculated that $$D$$ and $$D'$$ do not intersect.

• Note that your fractions can be simplified. $\frac{1}{-1} = -1$ and $\frac 1 1 = 1$; and thus $\frac{y-2}{-1} = 2-y$ and $\frac{y-3}{1} = y-3$.
– Stef
Oct 4 at 6:55
• @Stef The OP is of course aware of that, but chose a usual way to present a line:$$\frac{x-x_0}a=\frac{y-y_0}b=\frac{z-z_0}c.$$ Oct 4 at 9:00

A unit normal vector $$n=(a,b,c)$$ to that plane $$(\alpha)$$ must be orthogonal to $$v$$ and make an angle $$30^\circ$$ with $$v'.$$ The solutions of $$a^2+b^2+c^2=1,\quad a-b+c=0,\quad|2a+b-c|=\|v'\|\cos(30^\circ)$$ are $$n_1=\pm\frac1{\sqrt2}(1,1,0)\quad\text{and}\quad n_2=\pm\frac1{\sqrt2}(1,0,-1).$$ Using that $$(0,2,0)\in D\subset(\alpha),$$ we find two planes satisfying the constraints:

$$(\alpha_1):\quad x+y-2=0$$ and $$(\alpha_2):\quad x-z=0.$$

• thank you so much for your response. Oct 3 at 16:13
• @V'DollarTheppavongsa I hope it is correct. I have difficulties to visualize this double solution. Oct 3 at 16:16
• looking at the picture, the plane can either goes up from d to d' or go down from d to d' both cases make the angle 60 degrees. I didn't draw the possibility of α going down intersect d' at 60 degrees. I only draw where α goes up from d. Oct 3 at 16:27
• Even your drawing is hard to vizualize in 3D for me. Oct 3 at 16:29
• may I ask where you got a^2 +b^2+c^2 =1 Oct 3 at 16:41

The solutions given by Anne Bauval and Hosam H are very good.

I would like here to take a different approach (a little longer !) based on a general form of the planes passing through line $$D$$ (equation (2) below).

1. One can write the equations of $$D'$$ $${x-2\over2} = {y-3\over1} = {z+5\over-1}=\color{red}{t}$$ under the parametric form :

$$D' : \begin{cases}x&=&2\color{red}{t}+2\\y&=&\color{red}{t}+3\\z&=&-\color{red}{t}-5\end{cases}\tag{1}$$

1. Let us now find two points on line $$D$$, for example $$\pmatrix{0\\2\\0}$$ and $$\pmatrix{2\\0\\2}$$. Then, the general equation of planes passing through line $$D$$ and intersecting $$D'$$ is, under its determinantal form (see here) :

$$\left|\begin{matrix}0&2&\ \ 2t+2&x\\2&0&\ \ t+3&y\\0&2&-t-5&z\\1&1&1&1\end{matrix}\right|=0$$

Expanding, we obtain :

$$-4x+(-3t-7)y+(-3-3t)z+(6t+14)=0\tag{2}$$

with the alternative form ("pencil of planes") :

$$(-4x-7y-3z+14)+t(-3y-3z+6)=0\tag{2'}$$

From (3), we get a normal vector $$N=\pmatrix{-4\\(-3t-7)\\(-3-3t)}$$ which has to make a $$90°-60°=30°$$ angle with $$N'=\pmatrix{2\\1\\-1}$$, directing vector of line D'.

1. The $$30°$$ constraint can be transformed into an equation, by using the two ways one can express a dot product :

$$(N.N')^2=\|N\|^2\|N'\|^2 (\cos 30°)^2$$

$$144=(16+(-3t-7)^2+(3+3t)^2) \times 6 \times \frac34$$

giving a quadratic equation whose solutions are

$$t=-1 \ \ \text{and} \ \ t=-7/3.$$

"Plugging" into (1) these values of $$t$$ gives the equation of the two $$(\alpha)$$ planes :

$$x-z=0 \ \ \ \text{and} \ \ \ x+y-2=0\tag{3}$$

A remarkable fact is that these planes make a $$60°$$ between themselves (underlined in the solution by Hosam H). A comparison can be made with the following situation : a book is placed vertically on a table, opened with a 60° angle between its front and back cover (its binding features line $$D$$); now consider a line on the table (featuring line $$D'$$) completing an equilateral triangle with the bottom of the book. The line and the two planes are exactly in the same configuration as in the problem.

Fig. 1 : Line $$D$$ with points $$(0,2,0)$$ and $$(2,0,2)$$ (dark green) ; line $$D'$$ with points corresponding to parameters $$t=-1,-7/3$$ (yellow) : see equations (1) ; planes (blue and red) given by equations (2) and intermediate plane (grey) corresponding to intermediate value $$t=-5/3$$, these three planes belonging to the pencil of planes (2') "rotating" around line $$D$$.

• Excellent solution. [+1] Well-deserved. Oct 3 at 20:09
• Just corrected an error : $z=-t-5$ instead of $z=-t+5$ into parametric equations of $D'$ Oct 3 at 23:48

A plane $$\pi$$ containing line $$D$$ has its normal vector perpendicular to $$D$$.

The unit direction vector of $$D$$ is $$\omega =\dfrac{1}{\sqrt{3}}(1, -1, 1)$$. Two unit vectors that are perpendicular to $$\omega$$ are

$$u_1 = \dfrac{1}{\sqrt{2}} (1, 1, 0)$$

and

$$u_2 = \omega \times u_1 = \dfrac{1}{\sqrt{6}} (-1, 1, 2)$$

So now the unit normal vector $$n$$ of any plane containing line $$D$$, can be parameterized by $$\theta \in \mathbb{R}$$ as follows

$$n = \cos \theta u_1 + \sin \theta u_2$$

Since it is required that line $$D'$$ makes an angle of $$60^\circ$$ with the plane, then this means that $$D'$$ makes an angle of $$30^\circ$$ with $$n$$.

The unit direction vector along $$D'$$ is $$\mu = \dfrac{1}{\sqrt{6}}( 2, 1, -1)$$

Therefore, it is required that

$$\cos 30^\circ = \mu \cdot n$$

Substituting $$\mu$$ and $$n$$ and evaluating the dot product we get

$$\cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2} \cos \theta - \dfrac{1}{2} \sin \theta$$

The right hand side is just $$\cos( \theta + \dfrac{\pi}{6} )$$

So by inspection, the solutions are $$\theta = 0$$ and $$\theta = - \dfrac{\pi}{3}$$

Substituting these two values gives the two possible planes with the desired properties.

The two planes pass through $$P_0 = (0,2,0)$$ which is on line $$D$$.

The first normal is

$$n_1 = u_1 = \dfrac{1}{\sqrt{2}}( 1, 1, 0)$$

and the second normal is

$$n_2 = \dfrac{1}{2 \sqrt{2}} (2, 0, -2)$$

Hence the two possible planes are

$$x + y - 2 = 0$$

and

$$x - z = 0$$

• [+1] Good solution too ! Angle $\pi/3$ was "awaited" (see my remark with the image of the "book") Oct 3 at 18:54
• Thanks. However, I think you forgot to actually award me the [+1]. Oct 3 at 20:08
• Oops... Now done... Oct 3 at 20:33