TS EAMCET 4-Aug-2021 Shift 2 Question Paper

The equation of plane in cartesian form is given by
ax+by+cz+d=0 . . . (i)
‌ Normal vector to Eq. (i) is ‌2

∧

i

+6

∧

j

∴ Direction ratios to the normal of eq. (i) is 2,6,−3
∴a=2,b=6 and c=−3
and distance of plane from origin is 6
i.e. ‌

|a⋅0+b⋅0+c⋅0+d|

√a2+b2+c2

=6
⇒‌‌‌

|d|

√22+62+(−3)2

=6
⇒‌‌|d|=6×7=42
⇒‌‌d=±42
∴ Required plane 2x+6y−3z−42=0