KEAM 2014 Math Paper

Coordinate of mid-point of P(4,5,−10) and Q(−1,2,1) is
(‌

4−1

2

,‌

5+2

2

,‌

−10+1

2

) i.e.. (‌

3

2

,‌

7

2

,‌

−9

2

)
Now, DR's of PQ is (−1−4,2−5,1+10)
i.e.. (−5,−3,11) or (5,3,−11)
∴ Equation of plane passing through
(‌

3

2

,‌

7

2

,‌

−9

2

) and having DR's (5,3,−11) is
⇒‌‌5x+3y−11z=‌

135

2

It is written in vector form
r⋅(5

∧

i

+3

∧

j

−11

∧

k

)=‌

135

2