The foot of the perpendicular drawn from the point (1, 1. 1) to the plane π1 is /91, 3, 5).If (2,2, -1), (3, 4,2), (3, 3 ,0 ) are three points on the plane π2, then the angle between the planes π1 and π2 is
We know that, foot of perpendicular from a point (x1,y1,z1) on the plane ax+by+cz+d=0 is (x,y,z) where ‌
x−x1
a
=‌
y−y1
b
=‌
z−z1
c
=‌
−(ax1+by1+cz1+d)
a2+b2+c2
Since, the foot ofthe perpendicular drawn from the point (1,1,1) to the plane π1 is (1,3,5) ∴‌‌‌
1−1
a
=‌
3−1
b
=‌
5−1
c
=‌
−(a+b+c+d)
a2+b2+c2
⇒‌
0
a
=‌
2
b
=‌
4
c
=‌
−(a+b+c+d)
a2+b2+c2
⇒‌‌a=0,b=2k,c=4k Equation of plane π1 through point (1,3,5) and having directions of normal to plane are a,b,c is 0(x−1)+2k(y−3)+4k(z−5)=0 ⇒‌‌(y−3)+2(z−5)=0 ⇒y+2z−13=0 And equation of plane π2 is ⇒|
x−2
y−2
z+1
1
2
3
1
1
1
|=0 ⇒(x−2)(2−3)−(y−2)(1−3) ⇒ ⇒(x−2)+(y−2)(2)+(z+1)=0 ⇒x−2y+z−3=0 ∴ Angle between plane π1 and π2 θ=cos−1(|‌