Class 12 Maths Three Dimensional Geometry | Plane passing through intersection of 2 planes:Vector |

**Plane passing through intersection of 2 planes:Vector**

Let π_{1}and π_{2} are two planes with equations r^{->}. n̂_{1} = d_{1} and r^{->} .n̂_{2}=d_{2}

The position vector of any point on the line of intersection satisfies both the equations.

If t^{->} is the position vector of a point on the line, then

^{t->}. n̂_{1} = d_{1} and t^{->}.n̂_{2}=d_{2}

Therefore, for all real values of λ, we have

^{t->}.( n̂_{1 }+ λ n̂_{2})=d_{1}+ λd_{2}

Since t^{->} is arbitrary, it satisfies for any point on the line.

Hence the equation (r^{->}. (n_{1}⃗+ λn_{2}⃗))=d_{1}+ λd_{2} represents a plane π_{3 }which is such that if any vector r^{->} satisfies both the equations π_{1 }and π_{2}.

Then it will also satisfy the equation π_{3}e., any plane passing through the intersection of the planes.

Therefore r^{->}. n̂_{1} = d_{1} and r^{->}.n̂_{2}=d_{2} has the equation,

**r****⃗****.( n̂ _{1 }+ λ**

__Cartesian form:-__

Let n_{1}⃗= (A_{1}î+ B_{1}ĵ+ C_{1}k̂) and n_{2}⃗= (A_{2}î+ B_{2}ĵ+ C_{2}k̂)

r^{->}= xî+ y ĵ+ zk̂

Then equation(1) becomes,

x (A_{1 }+ λA_{2}) + y (B_{1} + λB_{2}) + z (C_{1} + λC_{2}) = d_{1} + λd_{2}

or **(A _{1} x + B_{1} y + C_{1}z – d_{1}) + λ(A_{2} x + B_{2} y + C_{2} z – d_{2}) = 0** ... (2)

This is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.

__Problem:-__

Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and

x + y + z – 2 = 0 and the point (2, 2, 1).

__Answer:-__

The equation of any plane through the intersection of the planes,

3x − y + 2z − 4 = 0 and x + y + z − 2 = 0, is

(3x − y + 2z – 4)+ λ (x + y + z – 2) = 0 where λ ∈ R. (1)

The plane passes through the point (2, 2, 1).

Therefore, this point will satisfy equation (1).

Therefore (3x2-2+2x1-4) + λ(2+2+1-2)=0.

=>2+3λ=0.

=> λ= (-2/3)

Substituting λ= (-2/3) in equation (1), we obtain

(3x − y + 2z – 4)(-2/3)(x + y + z – 2)=0

=>3(3x − y + 2z – 4)-2(x + y + z – 2) =0

=> (9x-3y+6z-12)-2(x+y+z-2)=0

=>7x-5y+4z-8=0.

This is the required equation of the plane.

__Problem:-__

Find the vector equation of the plane passing through the intersection of the planes r⃗. (î+ĵ+k̂)=6 and r⃗.(2 î+3ĵ+4k̂)=-5, and the point(1, 1, 1).

__Answer:-__

n_{1}⃗=(î+ ĵ+ k̂) and n_{2}⃗= (2 î+3ĵ+4k̂) andd_{1}=6, d_{2}=-5

Using the relation (r^{->}. (n_{1}⃗+ λn_{2}⃗))=d_{1}+ λd_{2}, we get,

r^{->}. [(î+ĵ+k̂) +λ (2 î+3ĵ+4k̂)]=6-5λ

Or r^{->}. (1+2 λ) x+ (1+3 λ) y + (1+4 λ) z=6-5 λ (1)

Where, λ is some real number.

Taking r^{->}= (xî+ yĵ+ zk̂), we get

(xî+ y ĵ+ zk̂). [(1+2 λ) î+ (1+3 λ) ĵ+ (1+4 λ) k̂]=6-5 λ

Or(1+2 λ)x+(1+3 λ)y + (1+4 λ) z=6-5 λ

Or(x+y+z-6) + λ (2x+3y+4z+5) =0 (2)

Given that the plane passes through the point (1, 1, 1), it must satisfy (2), i.e.

(1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0

Or λ = (3/14)

Putting the values of λ in (1), we get

r^{->} [((1+ (3/7)) î+ (1+ (9/14)) ĵ+ (1+ (6/7)) k̂)] = (6)-(15/14)

r^{->} ((10/7) î + (23/14) ĵ+ (13/14) k̂) = (69/14)

Or r^{->}. (20 î+23 ĵ+26 k̂)=69 which is the required vector equation of the plane.

__Problem:-__

Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and

2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

__Answer:-__

The equation of the plane through the intersection of the planes, x + y + z = 1 and

2x + 3y + 4z = 5, is

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

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

The direction ratios, a_{1}, b_{1}, c_{1}, of this plane are (2λ + 1), (3λ + 1), and (4λ + 1).

The plane in equation (1) is perpendicular to x – y + z=0.

Its direction ratios a_{2}, b_{2}, c_{2}, are 1, −1, and 1.

Since the planes are perpendicular,

a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0

=> (2λ + 1)-(3λ + 1) + (4λ + 1) =0

=>3λ + 1=0

=>λ= (-1/3)

Substituting λ= (-1/3) in equation (1), we obtain

This is the required equation of the plane.

(1/3) x-(1/3) z+92/3) =0

=>x-z+2=0.

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