There points
P=(1,3,2)Q=(−2,0,2)R=(1,4,3)
So, the vectorPQ and PR will be
PQ=(−2−1)^+(0−3)^+(2−2)k^=−3i^−3^PR=(1−1)i^+(4−3)^+(3−2)k^=^+k^
So, The vector perpendicular to this two vectors is given by
PQ×PR=∣∣i^−30j^−31k^01∣∣
So. The equation of plane will be
−3(x−1)+3(y−3)+(−3)(z−2)=0⇒x−y+z=2
Comments