JEE Main 24 Feb 2021 Shift 2 Solved Paper

Given, equation of line ⇒x−λ=2y−1=
⇒‌‌‌

x−λ

1

=‌

y−1∕2

1 2

=‌

z

−‌ 1 2

or ‌‌‌

x−λ

2

=‌

y−1∕2

1

=‌

z

−1

Point on this line through which it passes is (λ,‌

1

2

,0).
Equation of another line ⇒x=y+2λ=z−λ
⇒‌‌‌

x

1

=‌

y−(−2λ)

1

=‌

z−λ

1

... (ii)
A point through which this line passes is (0,−2λ,λ).
Now, distance between two skew lines
=‌

|(a2−a1)⋅(b1×b2)|

|b1×b2|

According to the question, ‌

|−5λ− 3 2 |

√14

=‌

√7

2√2

⇒‌‌|10λ+3|‌‌=7
10λ+3‌‌=±7
⇒‌‌10λ‌‌=4,−10
⇒‌‌λ‌‌=‌

2

5

‌ and ‌λ=−1
∴‌‌λ=−1
‌ ( ‌λ=‌

2

5

‌ is not possible as ‌λ‌ is an integer) ‌‌
∴‌
‌ Hence, ‌|λ|=|−1|=1‌