Class 12 Maths Three Dimensional Geometry Equation of a plane perpendicular to a given vector and pass

Equation of a plane perpendicular to a given vector and passing through a given point

There can be many planes that are perpendicular to the given vector, but through a given point P(x1,y1,z1) only one such plane exists.

Let a plane pass through a point A with position vector a-> and perpendicular to vector N->.

Let r-> be the position vector of any point P(x, y,z) in the plane.

Then the point P lies in the plane if and only if

AP-> is perpendicular to N->.i.e. AP->. N-> = 0.

But AP-> = r->. a->. Therefore (r->- a->). N->=0. ..(1)

This is the vector equation of the plane.

 

Cartesian form

Let the given point A (x1, y1, z1), P(x, y, z) and direction ratios of N⃗ are A,B and C.

Then a-> = x1î+y1ĵ+z1 k̂, r->=xî+yĵ+zk̂ and N->=Aî+Bĵ+C k̂

Now (r-> - a->). N-> =0

So [(x-x1) î +(y-y1) ĵ+ (z-z1) k̂].(Aî+Bĵ+C k̂)=0

e.A(x-x1)+B(y-y1)+C(z-z1) = 0

Problem:-

Find the vector and Cartesian equations of the plane which passes through the point (5, 2, – 4) and perpendicular to the line with direction ratios (2, 3, – 1).

Answer:-

We have the position vector of point (5, 2,-4) as a-> = 5î+2ĵ-4k̂ and the normal vector N-> perpendicular to the plane as N-> =2î+3ĵ-k̂.

Therefore, the vector equation of the plane is given by ( r->- a->). N-> =0

Or [r-> - (5î+2ĵ-4k̂)]. (2î+3ĵ-k̂)=0… (1)

Transforming (1) into Cartesian form, we have

[(x-5) î +(y-2) ĵ + (z+4) k̂]. (2î+3ĵ-k̂)= 0

Or 2(x-5) +3(y-2)-1(z+4) =0

i.e.  2x+3y-z=20, which is the Cartesian equation of the plane.

Problem:-

Find the vector and Cartesian equation of the planes

(a) That passes through the point (1, 0, −2) and the normal to the plane is î+ ĵ-k̂.

  (b) That passes through the point (1, 4, and 6) and the normal vector to the plane is î-2ĵ+ k̂.

Answer:-

The position vector of point (1, 0,-2) is a->= î-2k̂.

The normal vector N-> perpendicular to the plane is N->= î+ ĵ-k̂

The vector equation of the plane is given by, (r-> - a->). N-> =0

=>[r-> (î-2k̂)]. (î+ ĵ-k̂) =0 … (1)

r-> is the position vector of any point P(x,y,z) in the plane.

Therefore r-> =xî+yĵ+zk̂

Therefore equation (1) becomes

[(xî+yĵ+zk̂)-(î-2k̂)]. (î+ ĵ-k̂) =0

=> [(x-1) î + yĵ+ (z+2) k̂]. (î+ ĵ-k̂)=0

=>(x-1) +y-(z+2) =0

=>x+y-z-3=0

=>x+y-z=3.

This is the Cartesian equation of the required plane.

(a)  The position vector of the point (1, 4, 6) is a->= î+4ĵ+6k̂.

The normal vector N-> perpendicular to the plane is N-> = î-2ĵ+k̂

 The vector equation of the plane is given by, (r-> - a->). N-> =0

=>[r->- (î+4ĵ+6k̂)]. (î- 2ĵ +k̂) =0 … (1)

r-> is the position vector of any point P(x,y,z) in the plane.

Therefore r-> =xî+yĵ+zk̂

Therefore equation (1) becomes

[(xî+yĵ+zk̂)-(î+4ĵ+6k̂)]. (î- 2ĵ +k̂) =0

=> [(x-1)î + (y-4) ĵ+ (z-6)k̂].(î- 2 ĵ+k̂)=0

=>(x-1) -2(y-4) + (z-6) =0

=>x-2y+z+1=0.

This is the Cartesian equation of the plane.

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