MHT CET 2021 Math Solved Paper

The given equations are 3l+m+5n=0‌‌. . .(i)
and 6mn−2nl+5∕m=−0. . . (ii)
Now, from Eq. (i), we get
m=−3l−5n
On substituting m=−3∕−5n in Eq. (ii), we get
‌‌‌6(−3l−5n)n−2nl+5l(−3l−5n)=0
⇒‌30n2+45‌ln+15I2=0
⇒‌2n2+3‌ln+I2=0
⇒‌2n2+2nl+nl+I2=0
⇒‌2n(n+l)+l(n+l)=0
⇒‌‌‌(n+l)(2n+l)=0
⇒‌‌ Either ‌I=−n‌ or ‌I=−2n
‌ lf ‌I=−n,‌‌ then ‌m=−2n‌ and if ‌I=−2n‌, then ‌m=n
Thus, the direction ratios of two lines are proportional (−n,−2n,n) and (−2n,n,n) i.e. (−1,−2,1) and (−2,1,1), respectively.
Now, let θ be the acute angle between the lines,
‌ Then, ‌cos‌θ‌‌=‌

|a1a2+b1b2+c1c2|

√a12+b12+c12√a22+b22+c22

‌‌=‌

|2−2+1|

√1+4+1√4+1+1

=‌

1

6

⇒‌‌θ‌‌=cos−1(‌

1

6

)