JEE Main 26-Aug-2021 Shift 1 Solved Paper

Given line,

x−1

2

=

y−3

1

=

z−4

2

Given plane, x−2y−z=3
To find the projection let's find the foot of perpendicular from (1,3,, 4) to plane x−2y−z=3

x−1

1

=

y−3

−2

=

z−4

−1

=λ1
(λ1+1)−2(−2λ1+3)−(−λ1+4)=3
⇒6λ1=12
⇒ λ1=2
So, foot of perpendicular from (1,3,4) to plane x−2y−z=3 is A (3,−1,2).
Let us also find the intersection point of the plane and line

x−1

2

=

y−3

1

=

z−4

2

=λ2
(2λ2+1)−2(λ2+3)−(2λ2+4)=3−2λ2 =12
⇒ λ2=−6
The intersection point of the plane and line is B(−11,−3,−8)
Line passing through A and B is

x−3

−14

=

y+1

−2

=

z−2

−10

=μ

x−3

7

=

y+1

1

=

z−2

5

=μ
Now, let's find the distance from O(0,0,6) to this line L.
Let's say C(7μ+3,μ−1,5μ+2) is any point on L.
Then,
⇒ 49μ+21+μ−1+25μ−20=0
⇒ μ = 0
C(3,−1,2)
Distance =√(3−0)2+(−1−0)2+(2−6)2=√26
d2=26