Consider the planes 3x − 2y + z + 6 = 0 and 2x + y + 2z − 6 = 0 :
(a) Find the angle between the planes. (2 marks)
(b) Find the equation of the plane passing through the line of intersection of the above planes and through the point (0, 0, 0). (2 marks)

Question

Accepted Answer

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> [!TLDR]
> (a) The angle between two planes equals the angle between their normal vectors, found via the dot-product formula. (b) Use the family of planes through the line of intersection and pin down the constant using the given point.
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> **(a) Angle between $3x-2y+z+6=0$ and $2x+y+2z-6=0$**
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> Normals: $\vec n_1=(3,-2,1)$, $\vec n_2=(2,1,2)$.
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> $\vec n_1\cdot\vec n_2 = 3(2)+(-2)(1)+1(2) = 6-2+2 = 6$.
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> $|\vec n_1|=\sqrt{9+4+1}=\sqrt{14}$, $|\vec n_2|=\sqrt{4+1+4}=3$.
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> $\cos	heta = \dfrac{\vec n_1\cdot\vec n_2}{|\vec n_1||\vec n_2|} = \dfrac{6}{3\sqrt{14}} = \dfrac{2}{\sqrt{14}}$
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> $	heta = \cos^{-1}\left(\dfrac{2}{\sqrt{14}}ight) \approx 57.7^\circ$.
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> **(b) Plane through the line of intersection, passing through $(0,0,0)$**
> …

Q.Consider the planes 3x − 2y + z + 6 = 0 and 2x + y + 2z − 6 = 0 :

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

Q.Consider the planes 3x − 2y + z + 6 = 0 and 2x + y + 2z − 6 = 0 :

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