# Math Help - Scalar Equation of a Plane

1. ## Scalar Equation of a Plane

Please show me step by step solution!!!

Question 1:
Determine whether the following pairs of planes are coincident, parallel and distinct or neither.

a) $x + 3y - z - 2 = 0$ and $2x + 6y - 2z - 8 = 0$

I don't know how to put scalar equations into parametric equations...and how to find the direction vector...

-----------------

Question 2:
The angle between 2 planes is defined as the angle between their normals. Determine the angle (0 ≤ θ ≤ 90), to the nearest degree, between the given planes.

a) $2x + 3y - z + 9 = 0$ and $x + 2y + 4 = 0$

For #2, do you use the cross product?

sinθ = (|u x v|) / (|u||v|)

Again...for this question...I don't know how to put the scalar equation into parametric equation or find the direction vector..I know you use the algebraic cross product to find the normal vector, but I don't know what method to use to find the direction vector...I also know for a line, d = (d1, d2) and n = (-d2, d1)...

Thanks.

2. Originally Posted by Macleef
Please show me step by step solution!!!

Question 1:
Determine whether the following pairs of planes are coincident, parallel and distinct or neither.

a) $x + 3y - z - 2 = 0$ and $2x + 6y - 2z - 8 = 0$

I don't know how to put scalar equations into parametric equations...and how to find the direction vector...

-----------------

Question 2:
The angle between 2 planes is defined as the angle between their normals. Determine the angle (0 ≤ θ ≤ 90), to the nearest degree, between the given planes.

a) $2x + 3y - z + 9 = 0$ and $x + 2y + 4 = 0$

For #2, do you use the cross product?

sinθ = (|u x v|) / (|u||v|)

Again...for this question...I don't know how to put the scalar equation into parametric equation or find the direction vector..I know you use the algebraic cross product to find the normal vector, but I don't know what method to use to find the direction vector...I also know for a line, d = (d1, d2) and n = (-d2, d1)...

Thanks.
For the first one we need to find the normal vector to the plane.

if the plane is in the form $Ax+By+Cz+D=0$ then the normal vector is $Ai+Bj+Ck$

We also know that two planes are parallel iff their normal vectors are parallel.

So the two normal vectors are $i+3j-k$ and $2i+6j-2k$ these two vectors are parallel (the second is 2 times the first)

so the planes are parallel since their normal vectors are parallel.

The planes are distinct becuase they are not multiples of each other.

For the second I would use the Dot Product

the angle between two vectors is

$\theta=\cos^{-1}\left( \frac{u\cdot v}{|u||v|} \right)$

the normals are

$u=2i+3j-k \mbox{ and } v=i+2j+4k$

doing some computations

$u \cdot v = (2)(1)+(3)(2)+(-1)(4)=4$

$u=\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$

$v=\sqrt{1^2+2^2+4^2}=\sqrt{21}$

plugging into above we get

$\theta=\cos^{-1}\left( \frac{4}{\sqrt{14} \cdot \sqrt{21}}\right) \approx 76^{\circ}$

so the angle is $90^{\circ}-76^{\circ}=14^{\circ}$

P.s I think that 13.5 is a better answer