Given :
P = (√2, -1, 1)
To Find : Equation of plane
Formulae :
If \(\bar{n}\) = a\(\hat{i}\) + b\(\hat{j}\) + c\(\hat{k}\) is the vector normal to the plane, then equation of the plane is
Therefore direction cosines of the normal vector of the plane are (l, m, n)
Hence direction ratios are (kl, km, kn)
Therefore the equation of normal vector is
⇒ 4√2x + 4y + 4z = p
As point P (√2, -1, 1) lies on the plane by substituting it in above equation,
4√2(√2) + 4(-1) + 4(1) = p
⇒ 8 – 4 + 4 = p
⇒ P = 8
From eq(1)
This is the equation of required plane.