JEE 2023 Mathematics PYQ — Let the equation of the plane passing through the line of interse… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let the equation of the plane passing through the line of intersection of the planes x+2y+az=2 and x−y+z=3 be 5x−11y+bz=6a−1. For c∈Z, if the distance of this plane from the point (a,−c,c) is a2, then ca+b is equal to:
Choose the correct answer:
A.
-4
B.
2
C.
-2
D.
4
(Correct Answer)
Correct Answer:
4
Explanation
Solution
1. Concept of Family of Planes
The equation of any plane passing through the line of intersection of two planes P1=0 and P2=0 is P1+λP2=0.
(x+2y+az−2)+λ(x−y+z−3)=0
Grouping the variables together:
(1+λ)x+(2−λ)y+(a+λ)z−(2+3λ)=0
2. Comparison with the Given Equation
The problem provides the resulting plane as:
5x−11y+bz=6a−1
Since these two equations represent the same plane, their coefficients must be proportional:
51+λ=−112−λ=ba+λ=6a−12+3λ
From the first two parts, we find λ:
−11(1+λ)=5(2−λ)
−11−11λ=10−5λ⟹−6λ=21⟹λ=−27
3. Finding a and b
Substitute λ=−7/2 into the ratio 51+λ to find the scale factor:
Ratio=51−7/2=5−5/2=−21
Now, equate the constant terms to find a:
−21=6a−12+3(−7/2)⟹−21=6a−1−17/2
6a−1=17⟹6a=18⟹a=3
Next, find b using the z-coefficient:
−21=ba+λ⟹−21=b3−7/2=b−1/2⟹b=1
4. Finding c using the Distance Formula
The plane equation is 5x−11y+z−17=0. The point is (a,−c,c), which is (3,−c,c). The distance is a2=32.
52+(−11)2+12∣5(3)−11(−c)+1(c)−17∣=32
147∣15+11c+c−17∣=32⟹73∣12c−2∣=32
∣12c−2∣=14
Since c is an integer (c∈Z):
12c−2=−14⟹12c=−12⟹c=−1
Final Answer
Calculating the final expression:
ca+b=−13+1=−4
Explanation
Solution
1. Concept of Family of Planes
The equation of any plane passing through the line of intersection of two planes P1=0 and P2=0 is P1+λP2=0.
(x+2y+az−2)+λ(x−y+z−3)=0
Grouping the variables together:
(1+λ)x+(2−λ)y+(a+λ)z−(2+3λ)=0
2. Comparison with the Given Equation
The problem provides the resulting plane as:
5x−11y+bz=6a−1
Since these two equations represent the same plane, their coefficients must be proportional:
51+λ=−112−λ=ba+λ=6a−12+3λ
From the first two parts, we find λ:
−11(1+λ)=5(2−λ)
−11−11λ=10−5λ⟹−6λ=21⟹λ=−27
3. Finding a and b
Substitute λ=−7/2 into the ratio 51+λ to find the scale factor:
Ratio=51−7/2=5−5/2=−21
Now, equate the constant terms to find a:
−21=6a−12+3(−7/2)⟹−21=6a−1−17/2
6a−1=17⟹6a=18⟹a=3
Next, find b using the z-coefficient:
−21=ba+λ⟹−21=b3−7/2=b−1/2⟹b=1
4. Finding c using the Distance Formula
The plane equation is 5x−11y+z−17=0. The point is (a,−c,c), which is (3,−c,c). The distance is a2=32.
