Question

find the point of intersection of the planes x+y+z=6, 2x-y+3z=9 & x+3y+2z=13
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Answer 1

Answer:

Given planed

x+2y+3z=8 ...(1)

2x+3y+4z=11 ...(2)

and fine

1

x+1

=

2

y+1

=

3

z+1

...(3)

Let the plane passing through

intersection of plane (1) & (2)

(x+2y+3z−8)+k(2x+3y+4z−11)=0 ....(4)

Now, we have to show that line of inter.

(1) & (2) are co-plane with k=line (3)

if line (3) is co-planar with intersection of plane

Then, line (3) is point (−1,−1,−1)

passing through plane (4)

& normal to the plane (4) is 1st to 11 vector of line (3)

but (−1,−1,−1) in eqn.

(−1+2(−1)+3(−)−8)+k(2(−1)+3(−1)+4(−1)−11)=0

+14−20k=0

K=

10

−7

put K is (4)

(x+2y+3z−8)+(

10

−7

)(2x+3y+4z−11)=0

10x+10y+30z−80−14x−21y−28z+77=0

+4x−y+2z−3=0

4x+y−2z+3=0

Then this is eq

h

of plane

Direction ratio of normal of plane =4,1,−2

So, a

1

=4,b

1

=1,c

1

=−2

Direction ratios of line =1,2,3

so, a

2

=1,b

2

=2,c

2

=3

a

1

a

2

+b

1

b

2

+c

1

c

2

=4+2−6=0