Find the vector equation of the plane passing through the point (1, 1,

1.	Find the vector equation of the plane passing through the point (1, 1, 1) and parallel to the plane \(\bar{r}\).(2\(\hat{i}\) - \(\hat{j}\) + 2\(\hat{k}\)) = 5
Answer» Formula : Plane = r . (n) = d Where r = any random point n = normal vector of plane d = distance of plane from origin If two planes are parallel , then their normal vectors are same. Therefore , Parallel Plane r . (2i - j + 2k) = 5 Normal vector = (2i - j + 2k) ∴ Normal vector of required plane = (2i - j + 2k) Equation of required plane r . (2i - j + 2k) = d In cartesian form 2x - y + 2z = d Plane passes through point (1,1,1) therefore it will satisfy it. 2(1) - (1) + 2(1) = d d = 2 – 1 + 2 = 3 Equation of required plane r . (2i - j + 2k) = 3

Find the vector equation of the plane passing through the point (1, 1, 1) and parallel to the plane \(\bar{r}\).(2\(\hat{i}\) - \(\hat{j}\) + 2\(\hat{k}\)) = 5

Answer»

Formula : Plane = r . (n) = d

Where r = any random point

n = normal vector of plane

d = distance of plane from origin

If two planes are parallel , then their normal vectors are same.

Therefore ,

Parallel Plane r . (2i - j + 2k) = 5

Normal vector = (2i - j + 2k)

∴ Normal vector of required plane = (2i - j + 2k)

Equation of required plane r . (2i - j + 2k) = d

In cartesian form 2x - y + 2z = d

Plane passes through point (1,1,1) therefore it will satisfy it.

2(1) - (1) + 2(1) = d

d = 2 – 1 + 2 = 3

Equation of required plane r . (2i - j + 2k) = 3