JEE 2023 Mathematics PYQ — Let be the equation of a plane through the point and perpendicula… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let αx+βy+γz=1 be the equation of a plane through the point (3,−2,5) and perpendicular to the line joining the points (1,2,3) and (−2,3,5). Then the value of αβγ is equal to:
Choose the correct answer:
A.
6
(Correct Answer)
B.
201
C.
203
D.
101
Correct Answer:
6
Explanation
Solution
1. Find the Normal Vector to the Plane
The plane is perpendicular to the line joining points A(1,2,3) and B(−2,3,5). This means the vector AB is the normal vector(n) to the plane.
n=AB=(−2−1,3−2,5−3)
n=(−3,1,2)
2. Write the Equation of the Plane
The general equation of a plane with normal vector (a,b,c) passing through point (x1,y1,z1) is:
a(x−x1)+b(y−y1)+c(z−z1)=0
Using point (3,−2,5) and normal vector (−3,1,2):
−3(x−3)+1(y−(−2))+2(z−5)=0
−3x+9+y+2+2z−10=0
−3x+y+2z+1=0
3. Rearrange to the Given Form
The question gives the form αx+βy+γz=1. Let's move the constant to the right side:
−3x+y+2z=−1
To make the right side +1, multiply the entire equation by −1:
3x−y−2z=1
4. Identify the Coefficients
By comparing 3x−y−2z=1 with αx+βy+γz=1, we get:
α=3
β=−1
γ=−2
5. Calculate the Final Value
The question asks for the product αβγ:
αβγ=(3)×(−1)×(−2)
αβγ=6
Correct Option:(A) 6
Explanation
Solution
1. Find the Normal Vector to the Plane
The plane is perpendicular to the line joining points A(1,2,3) and B(−2,3,5). This means the vector AB is the normal vector(n) to the plane.
n=AB=(−2−1,3−2,5−3)
n=(−3,1,2)
2. Write the Equation of the Plane
The general equation of a plane with normal vector (a,b,c) passing through point (x1,y1,z1) is:
a(x−x1)+b(y−y1)+c(z−z1)=0
Using point (3,−2,5) and normal vector (−3,1,2):
−3(x−3)+1(y−(−2))+2(z−5)=0
−3x+9+y+2+2z−10=0
−3x+y+2z+1=0
3. Rearrange to the Given Form
The question gives the form αx+βy+γz=1. Let's move the constant to the right side:
−3x+y+2z=−1
To make the right side +1, multiply the entire equation by −1:
