1
Question
If z1,z2,z3 are three complex numbers such that 5z1−13z2+8z3=0, then the value of ∣∣
∣∣z1¯z11z2¯z21z3¯z31∣∣
∣∣ is
Open in App
Solution
5z1−13z2+8z3=0
⇒5z1+8z35+8=z2
⇒∣∣
∣∣z1¯z11z2¯z21z3¯z31∣∣
∣∣=z1(¯¯¯¯¯z2−¯¯¯¯¯z3)−¯¯¯¯¯z1(z2−z3)+1(z2¯¯¯¯¯z3−¯¯¯¯¯z2z3)
=z1[513¯¯¯¯¯z1+813¯¯¯¯¯z3−¯¯¯¯¯z3]−¯¯¯¯¯z1[513z1+813z3−z3] +[513z1¯¯¯¯¯z3+813z3¯¯¯¯¯z3−513¯¯¯¯¯z1z3−813¯¯¯¯¯z3z3]
=z1[513¯¯¯¯¯z1−513¯¯¯¯¯z3]−¯¯¯¯¯z1[513z1−513z3] +[513z1¯¯¯¯¯z3−513¯¯¯¯¯z1z3]
=0
⎡⎢
⎢
⎢⎣If z1,z2,z3 are collinear then conditionfor collinearity is ∣∣
∣∣z1¯z11z2¯z21z3¯z31∣∣
∣∣=0⎤⎥
⎥
⎥⎦
⇒5z1+8z35+8=z2
⇒∣∣ ∣∣z1¯z11z2¯z21z3¯z31∣∣ ∣∣=z1(¯¯¯¯¯z2−¯¯¯¯¯z3)−¯¯¯¯¯z1(z2−z3)+1(z2¯¯¯¯¯z3−¯¯¯¯¯z2z3)
=z1[513¯¯¯¯¯z1+813¯¯¯¯¯z3−¯¯¯¯¯z3]−¯¯¯¯¯z1[513z1+813z3−z3] +[513z1¯¯¯¯¯z3+813z3¯¯¯¯¯z3−513¯¯¯¯¯z1z3−813¯¯¯¯¯z3z3]
=z1[513¯¯¯¯¯z1−513¯¯¯¯¯z3]−¯¯¯¯¯z1[513z1−513z3] +[513z1¯¯¯¯¯z3−513¯¯¯¯¯z1z3]
=0
⎡⎢ ⎢ ⎢⎣If z1,z2,z3 are collinear then conditionfor collinearity is ∣∣ ∣∣z1¯z11z2¯z21z3¯z31∣∣ ∣∣=0⎤⎥ ⎥ ⎥⎦
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
