Finding the angle between a line and a plane

Question

Given that the equation of the line is:

$$ \mbox{P:}\quad \left\{\begin{array}{rcrcrcr} 3x & - & y & + & z & = & 6 \\ x & + & 2y & + & z & = &-3 \end{array}\right. $$

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$$ \mbox{and the plane is A:}\quad x + 2y + z = 5. $$

I believe you need to find the vector and use it to find the angle between the vector of the line and the normal vector of the plane.

I tried finding two points for the first equation but couldn't move further from there.

Answer 1

The vectors (3,-1,1) and (1,2,1) are the normals to the two planes that describe the line. Hence, the vector along the line will be the cross product of these.

Similarly, the plane's normal vector is (1,2,1) (which in your example contains the line, so the angle is zero).

But in general, you can fine the angle, $\theta$ between the line and normal to the plane via

$$\cos(\theta) = \frac{\vec{p} . \vec{l}}{(||\vec{p}||) (||\vec{l}||)}$$

Where $\vec{l}$ and $\vec{p}$ are the vectors describing the direction of the line and the normal of the plane respectively.

and then $\frac{\pi}{2}- \theta$ is the angle you're after.

Answer 2

let $$\vec{a}(a_x,a_y,a_z)$$ the given direction vector and $$Ax+By+Cz+D=0$$ the equation of the given plan e, then $$\sin(\phi)=\frac{|Aa_x+Ba_y+Ca_z|}{\sqrt{A^2+B^2+C^2}\cdot \sqrt{a_x^2+a_y^2+a_z^2}}$$

Answer 3

This equation can be simplified to a parametric line equation by the following computations. $$\mbox{P:}\quad \left\{\begin{array}{rcrcrcr} 3x & - & y & + & z & = & 6 \\ x & + & 2y & + & z & = &-3 \end{array}\right.$$

Ideally we want to have two variables in an equation where we express a line, so that we can have a point and a direction vector which can vary on some variable say $t$. To achieve this in your equation we would have to subtract the first one with the second one in the $P$ expression, this gives:

$$2x - 3y = 3$$

Now since we have only two variables we can substitue say $x = t$. Now we can express $y$ with respect to $x$ which would be. $$y = -1 + \frac{2t}{3}$$

Now we have both $x$ and $y$ in terms of $t$, so let us express $z$ in terms of $t$ using the first expression in the $P$.

$$z = 5 - \frac{7t}{3}$$.

Hooray! We have a parametric form which is: $$ \left\{\begin{array}{rcrcrcr} x &=& 0& + & t& \\ y &=& -1& + & \frac{2t}{3}& \\ z &=& 5& + & \frac{7t}{3}& \end{array}\right.$$

Now what about the direction vector? The direction vector is the part expressed with the variable $t$ as $t \in \mathbb{R}$. $$\textbf{d} = \left( 1,\frac{2}{3},\frac{7}{3}\right) $$ This direction vector difines the line from some specific point. Now to get the angle between the line and the given plane let us consider the dot product of the normal vector of the line given by $$\textbf{n} = (1,2,1) $$

Now computing the dot product $$ \cos(\theta)=\frac{\langle \mathbf{d}, \mathbf{n} \rangle}{\lVert \mathbf{d} \rVert \lVert \mathbf{n} \rVert}$$

Important to mention that one should consider the acute angle which is found by $\frac{\pi}{2} - \theta$

In this specific case I will leave out the details of plugging values in but the solution should be:

$$\theta = 0.812691015 \text{ rad} = 46.5637652 \text{ degrees}$$