Finding the Angle Using the Unit circle?

Question

Find the angle between the planes

$ 5(x − 1) − 3(y + 2) + 2z = 0,$

$ x + 3(y − 1) + 2(z + 4) = 0.$

I am not too sure how to start and solve this problem I know the formula $\cos \theta=|U_{n_1}*U_{n_2}|$ is involved. This leads me to conclude that somehow the unit circle is involved. Or are you supposed to complete the square. Any suggestions on how to solve this would help me greatly.

Answer 1

You need to identify the normal vectors $n_1 = (5,-3,2)^T$ and $n_2=(1,3,2)^T$. Now you can use the dot product $<n_1,n_2>=|n_1||n_2|\cos(\alpha)$.

Answer 2

the first plane $5x-3y + 2z = 11$ is orthogonal to $a = (5, -3, 2)^\top$ and the second is orthogonal to $(1, -3, 2)^\top.$ the cosine of the angle between the planes is $\frac{a^\top b}{\sqrt{(a^\top a) \space (b^\top b)}}$

Answer 3

Now we can define planes if we know some point that plane crosses and normal vector of plane. Now suppose we know some point $p_0$ coincide with plane. Now the equation of any line that has point $p_0$ and some random point $p_0$ has following form:

$(x-x_0)i + (y-y_0)j + (z-z_0)k$.

Then if this line lies in that plane, dot product of normal vector of plane and this line must be $0$. assuming that normal vector has the form ($Ai + Bj + Ck$) we can get general form for plane by finding dot product and equating it with $0$ (this because if two vectors are perpendicular dot product is zero), that is:

$A(x-x_0) + B(y-y_0) + C(z-z_0)$

Now check your plane formulas with that one. We get one normal vector for each plane, namely:

$n_1 = (5,-3,2)$ and $n_2=(1,3,2)$

Now the angle between planes is defined as, angle between their normal vectors. Now it comes to main point that, we can find angle $\theta$ in following way. Dot product of two vector is given as:

$n_1 * n_2 = |n_1||n_2|cos\theta$. Solve this for $cos\theta$

Then you will use $arccos\theta$ to find $\theta$