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(x_{1},y_{1},z_{1}) 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 (x_{1}, y_{1}, z_{1}), P(x, y, z) and direction ratios of N⃗ are A,B and C.

Then a^{->} = x_{1}î+y_{1}ĵ+z_{1} k̂, r^{->}=xî+yĵ+zk̂ and N^{->}=Aî+Bĵ+C k̂

Now (r^{->} - a^{->}). N^{->} =0

So **[(x-x _{1})**

e.A(x-x_{1})+B(y-y_{1})+C(z-z_{1}) = 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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