Given that the equation of the line is:

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

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

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.

  • $\begingroup$ Hint: The line is described as the intersection of two planes. What relationship does its direction vector have to the normals of these planes? $\endgroup$ – amd Apr 13 at 18:44

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.

  • $\begingroup$ How do you get the normal vector of a plane? $\endgroup$ – inoi Apr 13 at 18:56
  • $\begingroup$ Normal vector of a plane given by $ax+by+cz=d$ is $(a,b,c)$. $\endgroup$ – Rohit Pandey Apr 13 at 18:57
  • $\begingroup$ And if you have two planes, $a_1x+b_1y+c_1z=d_1$ and $a_2x+b_2y+c_2z=d_2$, then the vector of the line they describe is $(a_1,b_1,c_1)\times (a_2,b_2,c_2)$ where $\times$ is the cross product: mathsisfun.com/algebra/vectors-cross-product.html $\endgroup$ – Rohit Pandey Apr 13 at 19:01
  • $\begingroup$ How can you find the dot product of the vectors without using the cosine of the needed angle? It shouldn't work using this formula: A ⋅ B = ||A|| ||B|| cos θ $\endgroup$ – inoi Apr 13 at 19:08
  • $\begingroup$ Let $A=(a_1,b_1,c_1)$ and $B=(a_2,b_2,c_2)$. Then the dot product is: $(a_1a_2+b_1b_2+c_1c_2)$. This is equivalent to the formula with the cosine, so you can equate the two to get the angle between them, $\theta$. $\endgroup$ – Rohit Pandey Apr 13 at 19:09

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 $$\sin(\phi)=\frac{|Aa_x+Ba_y+Ca_z|}{\sqrt{A^2+B^2+C^2}\cdot \sqrt{a_x^2+a_y^2+a_z^2}}$$

  • $\begingroup$ But how do I find the direction vector of the line and of the plane? $\endgroup$ – inoi Apr 13 at 18:55
  • $\begingroup$ For the plane are these the coefficients $$(1,2,1)$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 13 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.