We have to find the angle between the planes 2x−y+z=1 and x+y+2z=3 So, a1=2,b1=−1,c1=1 and a2=1,b2=1 and c2=2 Let θ will be angle between these two plane So, cosθ=
a1a2+b1b2+c1c2
√(a1)2+(b1)2+(c1)2√(a2)2+(b2)2+(c2)2
=
2×1+(−1)×1+1×2
√22+(−1)2+12√12+12+22
⇒cosθ=
2−1+2
√6√6
=
3
6
=
1
2
=cos
π
3
∴θ=π∕3 So, statement 1 is true. Now, we have to the find distance between the planes 6x−3y+6z+2=0 and 2x−y+2z+4=0 Rewrite the equation 6x−3y+6z+2=0 ⇒3(2x−y+2z+2∕3)=0 ⇒2x−y+2z+2∕3=0 So, 2x−y+2z+2∕3=0 and 2x−y+2z+4=0 are parallel a=2,b=−1,c=2,d1=2∕3 and d2=4 Distance =|