MHT CET 2017 Math Solved Paper

(c) Let n1 and n2 are normals to the planes
r. (m

^

i

−

^

j

+2

^

k

) + 3 = 0 and
r . (2

^

i

−m

^

j

−

^

k

) - 5 = 0 respectively.
Now, θ =

π

3

is angle between the planes.
As, cos θ = |

n1⋅n2

|n1|⋅|n2|

|
So,cos

π

3

=
⇒

1

2

= = |

3m−2

√(m2+5)2

|
⇒ ±

1

2

=

3m−2

(m2+5)2

m2 + 5 = 6m - 4 or -m2 - 5 = 6m - 4
⇒ m2 - 6m -9 = 0 or m2 + 6m + 1 = 0
⇒ (m-3) 2 = 0 or m2 + 6m + 1 = 0
As m2 + 6m + 1 = 0 does not give any real values.
Hence, (m−3)2 = 0 ⇒ m = 3