Q.Find the angle between the planes $x+2y+3z=6$ and $3x-3y+z=1$.

The angle between the planes is $\cos^{-1}\left(\frac{4}{21}\right)$.

The angle between two planes equals the angle between their normals. Normals are $\vec{n_1} = (2, 1, -2)$ and $\vec{n_2} = (3, -6, -2)$.

$\vec{n_1}\cdot\vec{n_2} = (2)(3) + (1)(-6) + (-2)(-2) = 6 - 6 + 4 = 4.$

$|\vec{n_1}| = \sqrt{4+1+4} = 3,\qquad |\vec{n_2}| = \sqrt{9+36+4} = 7.$ …

The angle between planes equals the angle between their normal vectors.

Plane 1: $2x+y+3z=2$, normal $\vec n_1=(2,1,3)$.

Plane 2: $x-2y+0z=5$, normal $\vec n_2=(1,-2,0)$.

$\cos\theta=\dfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}$

$\vec n_1\cdot\vec n_2=2(1)+1(-2)+3(0)=2-2+0=0$

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The angle between two planes equals the angle between their normal vectors; use $\cos\theta = \dfrac{\vec n_1\cdot\vec n_2}{|\vec n_1||\vec n_2|}$.

The planes are $2x-y+2z=3$ and $6x-2y+3z=5$, giving normal vectors $\vec n_1=(2,-1,2)$ and $\vec n_2=(6,-2,3)$.

$\vec n_1\cdot\vec n_2 = 2(6)+(-1)(-2)+2(3) = 12+2+6=20.$ …

The normal vectors of the two planes are scalar multiples of each other, so the planes are parallel (and distinct).

Plane 1: $2x-y+4z=5$, normal $\vec n_1=(2,-1,4)$.

Plane 2: $5x-2.5y+10z=6$, normal $\vec n_2=(5,-2.5,10)$.

Check if $\vec n_2$ is a scalar multiple of $\vec n_1$:

$\frac{5}{2}=2.5,\quad \frac{-2.5}{-1}=2.5,\quad \frac{10}{4}=2.5.$

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Angle between planes = angle between their normal vectors: $\cos\theta = \dfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}$.

Normals: $\vec n_1 = (1,1,2)$ (from $x+y+2z=3$), $\vec n_2=(3,-2,2)$ (from $3x-2y+2z=7$).

$\vec n_1\cdot\vec n_2 = 1(3)+1(-2)+2(2) = 3-2+4 = 5$

$|\vec n_1| = \sqrt{1+1+4}=\sqrt6$, $\quad |\vec n_2|=\sqrt{9+4+4}=\sqrt{17}$

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The angle between two planes equals the angle between their normal vectors, found via the dot product formula.

Normal to $x-y+2z=9$: $\vec n_1=(1,-1,2)$. Normal to $2x+y+z=7$: $\vec n_2=(2,1,1)$.

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