A vector of magnitude 8 units is inclined to the x-axis at 45^o, y-axis at 60^o and an acute angle with the z-axis, if a plane passes through a point (√2, -1, 1) and is normal to find its equation in vector form.

Given :

P = (√2, -1, 1)

To Find : Equation of plane

Formulae :

1)

Where

2) Equation of plane :

If is the vector normal to the plane, then equation of the plane is

As

and

But,

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

Now, equation of the plane is

………eq(1)

But

⇒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)

Dividing throughout by 4

This is the equation of required plane.