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Question

If the equation of the plane passing through the mirror image of a point (2,3,1)with respect to line (x+1)2=(y3)1=(z+2)-1 and containing the line (x2)3=(1y)2=(z+1)1 is αx+βy+z=24, then, α+β+γ is equal to:


A

21

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B

19

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C

18

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D

20

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Solution

The correct option is B

19


Step 1: Solve for the mirror image of the point with respect to the line:

JEE Main 2021 March 17 Section 1 Shift 2 Maths Question Paper with Solutions Q5

Given A(2,3,1)

Let the line (x+1)2=(y3)1=(z+2)-1be L1

We know that if a line is of the form x-x1p=y-y1q=z-z1r then

x1,y1,z1 denotes a point on the line and p,q,r denotes the direction cosines.

Let Ba,b,c is the mirror image of Awith respect to the line L1

Let (x+1)2=(y3)1=(z+2)-1=λ

Then any point on the line L1 is of the form (2λ1,λ+3,λ2)

Let point M be the mid point between the point (2,3,1) and its image with respect to the line

Since any point in the line is of the form (2λ1,λ+3,λ2)

Therefore, let M(2λ1,λ+3,λ2)

Direction ratio's of the line AM are: 2λ12,λ+33,λ21

=2λ3,λ,λ3

We know that AML1

Since perpendicular, there the product of the direction ratio will be zero.

2(2λ3)+1(λ)1(λ3)=06λ=3λ=12

M=(0,72,52)

M is mid-point of A&B

0,72,-52=a+22,b+32,c+12

a,b,c=-2,4,-6

Therefore the co-ordinate of B is -2,4,-6

Step 2: Solve for the equation of the plane

The plane pass through the point B and contain the line

(x2)3=(1y)2=(z+1)1

(x2)3=(y-1)-2=(z+1)1.........i

Let the (x2)3=(y-1)-2=(z+1)1 is L2

JEE Main 2021 March 17 Section 1 Shift 2 Maths Question Paper with Solutions Q5

We know that if the equation of the line is x-αa=y-βb=z-γcthe the line passes through α,β,γ and the direction cosines are a,b,c

Point P on line is(2,1,1)

The direction cosineb of the line of lineL2 is 3,2,1

PB=-2-2i^+4-1j^+-6+1k^=4i^+3j^5k^ B=-2,4,-6

We know equation of the plane passing the lines x-αa=y-βb=z-γc and a1i^+b1j^+c1k^ is

x-αy-βz-γabca1b1c1=0x-2y-1z+13-214-35=0x-2-10+3+y-14-15+z+1-9+8=0-7x+14-11y+11-z-1=07x+11y+z=24

Step 3: Solve for the required values

Now comparing with the equation αx+βy+z=24 we have

α=7β=11γ=1α+β+γ=19

Hence, option(B) i.e. 19 is correct


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