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The equation of the plane passing through the intersection of the planes  and and the point (1, 2, 1) is:​
  • a)
    18x+6y+14z-23=0
  • b)
    18x+7y+14z-46=0
  • c)
    9x+3y+7z-23=0
  • d)
    18x+7y+14z-38=0
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The equation of the plane passing through the intersection of the plan...
n1 = 2i + j + k
n2 = 2i + 3j - 4k
p1 = 4,   p2 = -6
r.(n1 + λn2) = p1 + λp2
=> r . [2i + j + k + λ(2i + 3j - 4k)] = 4 - 6λ
=> r . [ i(2 + 2λ) + j(1 + 3λ) + k(1 - 4k)] = 4 - 6λ
Taking r = xi + yj + zk
(2 + 2λ)x + (1 + 3λ)y + (1 - 4k)z = 4 - 6λ
(2x + y + - z - 4) + λ(2x + 3y - 4k + 6) = 0
Given points are (1,2,1) 
(2 + 2 - 1 - 4) + λ(2 + 6 - 4 + 6) = 0
-1 + λ(10) = 0
 λ = 1/10
Substitute  λ = 1/10, we get
18x + 7y + 14z - 46=0
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Most Upvoted Answer
The equation of the plane passing through the intersection of the plan...
I am giving you an example by that way will be easily be able to solve the question:
Q:
 
Find the equation of the plane that passes through the intersection of the planes 
r
⃗ 
.
(
i
^
+
j
⃗ 
+
k
⃗ 
)
=
6
 and 
r
⃗ 
.
(
2
i
^
+
3
j
⃗ 
+
4
k
⃗ 
)
=
5
 and passing through the point (1,1,1).
Answer: You must note that the vector equations of the two planes are given so the equation of the required (third) plane will also be in vector form.
Given:
π
1
=
r
⃗ 
.
(
i
^
+
j
⃗ 
+
k
⃗ 
)
=
6
π
2
=
r
⃗ 
.
(
2
i
^
+
3
j
⃗ 
+
4
k
⃗ 
)
=
5
Now, we already know that the equation of the required plane is 
π
1
+
λ
π
2
=
0
 i.e.
[
r
⃗ 
.
(
i
^
+
j
⃗ 
+
k
⃗ 
)
6
]
+
λ
[
r
⃗ 
.
(
2
i
^
+
3
j
⃗ 
+
4
k
⃗ 
5
]
=
0
r
⃗ 
.
[
(
i
^
+
j
^
+
k
^
)
+
λ
(
2
i
^
+
3
j
⃗ 
+
4
k
⃗ 
)
6
+
5
λ
=
0
(
x
i
^
+
y
j
^
+
z
k
^
)
.
[
(
i
^
+
j
^
+
k
^
)
+
λ
(
2
i
^
+
3
j
⃗ 
+
4
k
⃗ 
)
6
+
5
λ
=
0
Now since the required plane passes through (1,1,1), the point must satisfy the equation of the plane. Putting x =1 , y = 1, z = 1 we have –
(
i
^
+
j
^
+
k
^
)
.
[
(
i
^
+
j
^
+
k
^
)
+
λ
(
2
i
^
+
3
j
⃗ 
+
4
k
⃗ 
)
6
+
5
λ
=
0
(
i
^
+
j
^
+
k
^
)
.
[
(
1
+
2
λ
)
i
^
+
(
1
+
3
λ
)
j
^
+
(
1
+
4
λ
)
k
^
]
6
+
5
λ
=
0
Taking the dot product, we have:
1
+
2
λ
+
1
+
3
λ
+
1
+
4
λ
6
+
5
λ
=
0
λ
=
3
14
Replacing the value of \lambda in the equation of the required plane gives us –
r
⃗ 
.
[
(
i
^
+
j
^
+
k
^
)
+
3
14
(
2
i
^
+
3
j
⃗ 
+
4
k
⃗ 
)
6
+
5
3
14
=
0
r
⃗ 
.
[
(
1
+
2.
3
14
)
i
^
+
(
1
+
3
3
14
)
j
^
+
(
1
+
4
3
14
)
k
^
6
+
5
3
14
=
0
r
⃗ 
.
[
20
14
i
^
+
23
14
j
^
+
26
14
k
^
]
69
14
=
0
or,
r
⃗ 
.
(
20
i
⃗ 
+
23
j
⃗ 
+
23
k
⃗ 
)
=
69
is the required equation of the plane in Vector form.
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Community Answer
The equation of the plane passing through the intersection of the plan...
Equation of given planes in Cartesian form are
2x+y+z-4=0 and 2x+3y-4z+6=0
plane passing through intersection of these planes
2x+y+z-4+μ(2x+3y-4z+6)=0
point(1,2,1) lies on above plane so, substitute points in above plane in the place of x, y and z we get
μ=-1/10 substitute this value in the same equation
we get
18x+7y+14z-46=0
which is same as option B
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The equation of the plane passing through the intersection of the planesandand the point (1, 2, 1) is:​a)18x+6y+14z-23=0b)18x+7y+14z-46=0c)9x+3y+7z-23=0d)18x+7y+14z-38=0Correct answer is option 'B'. Can you explain this answer?
Question Description
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