If z 1- z 2- z 3 represent the vertices of an equilateral - ABC and - z 2- z 1-2- z 3- z 1-2- z 2- z 3-2 z 1-2- z 3- z 2-2-sin- where 0- then - isA- -4B- 2 -3C- -6D- 5 -6

The correct option is A π/4Δ ABC is an equilateral ⇒ |z_2 z_1| = |z_3 z_1| = |z_2 z_2| = a and length of median =√(3)/2a = | z_2+z_3/2 z_1 | ∴ sin θ = √(a^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

If z 1, z 2, ...

Question

If $z_{1}, z_{2}, z_{3}$ represent the vertices of an equilateral $Δ A B C$ and $\sqrt{\frac{| z_{2} - z_{1} |^{2} + | z_{3} - z_{1} |^{2}}{| z_{2} + z_{3} - 2 z_{1} |^{2} + | z_{3} - z_{2} |^{2}}} = s i n θ$ (where $0 < θ < π$ ) then $θ$ is

\frac{π}{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2 π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{5 π}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{π}{4}$
$Δ A B C$ is an equilateral $\Rightarrow | z_{2} - z_{1} | = | z_{3} - z_{1} | = | z_{2} - z_{2} | = a$ and length of median $= \frac{\sqrt{3}}{2} a = ∣ ∣ \frac{z_{2} + z_{3}}{2} - z_{1} ∣ ∣$
$∴ s i n θ = \sqrt{\frac{a^{2} + a^{2}}{3 a^{2} + a^{2}}} = \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4}$

Suggest Corrections

Similar questions

Q. If

z_{1}, z_{2}, z_{3}

represent the vertices of an equilateral triangle such that

| z_{1} | = | z_{2} | = | z_{3} |

, then

Q. Statement 1: If

| z_{1} | = | z_{2} | = | z_{3} |, z_{1} + z_{2} + z_{3} = 0

and

(z_{1}), B (z_{2}), C (z_{3})

are the vertices of

Δ A B C

, then one of the values of

a r g (\frac{(z_{2} + z_{3} - 2 z_{1})}{(z_{3} - z_{2})})

\frac{π}{2}

.
Statement 2: In equilateral triangle, orthocenter coincides with centroid.

Q. If the complex numbers,

z_{1}, z_{2}, z_{3}

represent the vertices of an equilateral triangle such that

| z_{1} | = | z_{2} | = | z_{3} |,

then

z_{1} + z_{2} + z_{3} =

Q. If

z_{1} + z_{2} + z_{3} = 0

, then

| z_{2} - z_{3} |^{2} + | z_{3} - z_{1} |^{2} + | z_{1} - z_{2} |^{2}

equals

Q. If the complex numbers

z_{1}, z_{2}, z_{3}

represent the the vertices of an equilateral triangle such that

| z_{1} | = | z_{2} | = | z_{3} |

, then prove that

z_{1} + z_{2} + z_{3} = 0

Join BYJU'S Learning Program

If z1,z2,z3 represent the vertices of an equilateral ΔABC and √|z2−z1|2+|z3−z1|2|z2+z3−2z1|2+|z3−z2|2=sin θ (where 0<θ<π) then θ is

The correct option is A π4ΔABC is an equilateral ⇒|z2−z1|=|z3−z1|=|z2−z2|=a and length of median =√32a=∣∣z2+z32−z1∣∣ ∴sin θ=√a2+a23a2+a2=1√2⇒θ=π4

If $z_{1}, z_{2}, z_{3}$ represent the vertices of an equilateral $Δ A B C$ and $\sqrt{\frac{| z_{2} - z_{1} |^{2} + | z_{3} - z_{1} |^{2}}{| z_{2} + z_{3} - 2 z_{1} |^{2} + | z_{3} - z_{2} |^{2}}} = s i n θ$ (where $0 < θ < π$ ) then $θ$ is