Find the equation of a plane (in a x + b y + c z = d form) which is at a distance of 3 3 units from the origin and the normal to which is equally inclined to the coordinate axes.

Given that the normal to the plane is equally inclined to the coordinate axis. Now if α , β , and γ are the angles made by the line segment with the coordinate axis. Then, we have cos α = cos β = cos γ We know that cos 2 α + cos 2 β + cos 2 γ = 1 ⇒ 3 cos 2 α = 1 ⇒ cos α = 1 3 ∵ α is an acute angle So, the normal to the plane is N → = 1 3 i ^ + 1 3 j ^ + 1 3 k ^ and plane is at a distance of 3 3 units from origin. We know that the equation of the plane at a distance p from the origin where N ^ is the unit vector normal to the plane is r → · N ^ = p . . . . . 1 Now, N → = 1 ( 3 ) 2 + 1 ( 3 ) 2 + 1 ( 3 ) 2 = 3 3 = 1 ⇒ N ^ = N → N → = 1 3 i ^ + 1 3 ȷ ^ + 1 3 k ^ 1 = 1 3 i ^ + 1 3 ȷ ^ + 1 3 k ^ Therefore, the equation of the required plane using 1 is ⇒ x i ^ + y j ^ + z k ^ · 1 3 i ^ + 1 3 ȷ ^ + 1 3 k ^ 1 = 3 3 ⇒ x 3 + y 3 + z 3 = 3 3 ⇒ x + y + z = 3 3 · 3 = 9 So, the required equation of plane is x + y + z = 9 .

Find the vector equation of the plane which is at a distance of 6 29 from the origin and its normal vector from the origin is 2 i ^ - 3 j ^ + 4 k ^ . Also, find its Cartesian form.

We know that the vector equation of a plane at a distance ' p ' from the origin and normal to the unit vector n ^ is r → · n ^ = p . Given that vector normal to the plane is n → = 2 i ^ - 3 j ^ + 4 k ^ . ∴ The unit vector normal to the plane is n ^ = 2 2 2 + - 3 2 + 4 2 i ^ - 3 2 2 + - 3 2 + 4 2 j ^ + 4 2 2 + - 3 2 + 4 2 k ^ ⇒ n ^ = 2 29 i ^ - 3 29 j ^ + 4 29 k ^ Here, given that p = 6 29 . Thus, the vector equation of the plane is r → · 2 29 i ^ - 3 29 j ^ + 4 29 k ^ = 6 29 ∴ The Cartesian equation of the plane is ( x i ^ + y j ^ + z k ^ ) · 2 29 i ^ - 3 29 j ^ + 4 29 k ^ = 6 29 ⇒ 2 x 29 - 3 y 29 + 4 z 29 = 6 29 ⇒ 2 x - 3 y + 4 z 29 = 6 29 ⇒ 2 x - 3 y + 4 z = 6 .