Answer :

**Given:** and is equally inclined with an acute with the coordinate axes

**To find:** the vector and Cartesian forms of the equation of a plane which passes through (2, 1, -1) and is normal to

Let has direction cosines as l, m and n and it makes an angle of α, β and γ with the coordinate axes. So as per the given condition

α=β=γ

⇒ cos α =cos β =cos γ

⇒ l=m=n=p (let assume)

We know that,

l^{2}+m^{2}+n^{2}=1

⇒ p^{2}+p^{2}+p^{2}=1

⇒ 3p^{2}=1

So,

For the negative value of cos the angles are obtuse so that we will neglect it

So we have

Hence

So the vector equation of the normal becomes,

The plane is passing through the point (2, 1, -1). Let the position vector of this point be

We know that vector equation of a plane passing through point and perpendicular/normal to the vector is given by

Substituting the values from eqn(i) and eqn(ii) in the above equation, we get

(by multiplying the two vectors using the formula )

is the vector and Cartesian forms of the equation of a plane which passes through (2, 1, -1) and is normal to .

Let

Then, the above vector equation of the plane becomes,

Now multiplying the two vectors using the formula, we get

This is the Cartesian form of the equation of a plane which passes through (2, 1, -1) and is normal to.

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