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

Plane passing through intersection of 2 planes:Vector

Let π1and π2 are two planes with equations r->. n̂1 = d1 and r-> .n̂2=d2

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->. n̂1 = d1 and t->.n̂2=d2

Therefore, for all real values of λ, we have

t->.( n̂1 + λ n̂2)=d1+ λd2

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

Hence the equation (r->. (n1⃗+ λn2⃗))=d1+ λd2 represents a plane π3 which is such that if any vector r-> satisfies both the equations π1 and π2.

Then it will also satisfy the equation π3e., any plane passing through the intersection of the planes.

Therefore r->. n̂1 = d1 and r->.n̂2=d2 has the equation,

r.( n̂1 + λ2)=d1+ λd2 (1)

Cartesian form:-

Let n1⃗= (A1î+ B1ĵ+ C1k̂) and n2⃗= (A2î+ B2ĵ+ C2k̂)

r->= xî+ y ĵ+ zk̂

Then equation(1) becomes,

x (A1 + λA2) + y (B1 + λB2) + z (C1 + λC2) = d1 + λd2

or (A1 x + B1 y + C1z – d1) + λ(A2 x + B2 y + C2 z – d2) = 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:-

n1⃗=(î+ ĵ+ k̂) and n2⃗= (2 î+3ĵ+4k̂) andd1=6, d2=-5

Using the relation (r->. (n1⃗+ λn2⃗))=d1+ λd2, 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, a1, b1, c1, 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 a2, b2, c2, are 1, −1, and 1.

Since the planes are perpendicular,

a1a2+b1b2+c1c2=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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