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The angle between the planes x+y+2z=6 and 2x-y+z=7 is: a)Pi/2 b)Pi/3 c)Pi/4 d)Pi/6 Plz explain..

The angle between the planes x+y+2z=6 and 2x-y+z=7 is:


a)Pi/2


b)Pi/3


c)Pi/4


d)Pi/6


Plz explain..

Grade:12

1 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

The form Ax + By + Cz = D is particularly useful because we can arrange things so that D gives the perpendicular distance from the origin to the plane.
To get this nice result, we need to work with the unit normal vector. This is the vector of unit length which is normal to the surface of the plane. (There are two choices here, depending on which direction you choose, but one is just minus the other).
I'll call this unit normal vector n.
Next we see how using n will give us D, the perpendicular distance from the origin to the plane.
In the picture below, P is any point in the plane. It has position vector r from the origin O.
from any r we can find D Now we work out the dot product of r and n. This gives us r.n = |r||n|cos A.
But |n| = 1 so we have r.n = |r|cos A = D.This will be true wherever P lies in the plane.

Next, we split both r and n into their components.
We write r = xi + yj + zk and n = n1i + n2j + n3k.
Therefore r.n = (xi + yj + zk) . (n1i + n2j + n3k) = D
so r.n = xn1 + yn2 + zn3 = D.
We see that n1, n2 and n3 (the components of the unit surface normal vector) give us the A, B and C in the equation Ax + By + Cz = D.

let the components of the unit surface normal vectors to the two planes be (n1,n2,n3) and (m1,m2,m3),

taking the dot product of the two vectors, we get (m1n1+m2n2+m3n3)=√(m1^2+m2^2+m3^2)*√(n1^2+n2^2+n3^2)cosθ, where theta is the angle between the two normals, and hence the angle between the two planes.

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