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I have the line: $$r:\\3z-x = 1\\y-1 = 1$$

And the plane makes an angle $\theta = \arccos \frac{2\sqrt{30}}{11}$ with the line: $$s:X = (1,1,0) + \lambda(3,1,1)$$

What I tried:

From the equations of $r$ we have $y=2$ and by letting $x=\lambda$ we get:

$$x = \lambda, y = 2, z = \frac{1}{3}+\frac{\lambda}{3}$$ So we already have one direction vector of the plane: $(1,0,\frac{1}{3})$ or $(3,0,1)$. By assuming the other to be $(a,b,c)$ then the normal to the plane is $$(3,0,1)\times(a,b,c) = (-b, a-3c, 3b)$$

We also know that the plane makes an angle $\theta = \arccos \frac{2\sqrt{30}}{11}$ with line $s$. By the formula of the angle between line and plane:

$$\sin\theta = \frac{|\vec v \cdot \vec n|}{||\vec v||||\vec n||}$$

We have, then:

$$\sin \arccos \frac{2\sqrt{30}}{11} = \frac{|(3,1,1)\cdot (-b, a-3c, 3b)|}{\sqrt{11}\sqrt{(-b)^2 + (a-3c)^2 + (3b)^2}}$$ ($\vec v$ is the direction vector of $s$)

But I can't solve it from here because I have 3 unknown variables :(

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1 Answer 1

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Let's just think about direction vectors, ignoring offsets for the moment. The condition that a given line lies in the plane constrains the normal vector of that plane to lie in a certain plane through the origin. The condition of the given angle can be satisfied by normal vectors which lie in a given cone which has its apex at the origin. So to satisfy both, you need to intersect a plane through the origin with a cone with its apex at the origin, but you must not choose the null vector as the solution. In general, such an intersection will be a pair of lines. This tells you that it should be all right if you choose one coordinate arbitrarily, and that you should expect a quadratic equation to give you these two results.

You have $\cos\theta=\frac{2\sqrt{30}}{11}$ from which you can conclude $\sin\theta=\pm\frac1{11}$. I'd choose $(a,b,c)$ directly as the normal, to avoid the cross product. Then you have these conditions:

\begin{align*} (3,0,1)\cdot(a,b,c) &= 0 & \frac{(3,1,1)\cdot(a,b,c)}{\sqrt{3^2+1^2+1^2}\sqrt{a^2+b^2+c^2}} &= \sin\theta=\pm\frac1{11} \\ 3a+c &= 0 & \sqrt{11}\,\lvert 3a+b+c\rvert &= \sqrt{a^2+b^2+c^2} \end{align*}

Now choose $a=1$ arbitrarily, to obtain a unique solution (i.e. to fix the length and sign of the normal vector).

\begin{align*} a &= 1 & 3a + c &= 0 & \sqrt{11}\,\lvert 3a+b+c\rvert &= \sqrt{a^2+b^2+c^2} \\ && 3 + c &= 0 & \sqrt{11}\,\lvert b\rvert &= \sqrt{1+b^2+9} \\ && c &= -3 & 11b^2 &= 10+b^2 \\ && && 10b^2 &= 10 \\ && && b&=\pm1 \end{align*}

So you have two possible normal vectors, namely $\vec n_1=(1,1,-3)$ and $\vec n_2=(1,-1,-3)$.

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