Az1- Bz2- Cz3 are the vertices of the triangle ABC in anticlockwise order- If - ABC - -4 and AB - -2BC-Then prove that z2 - z3 - iz1 - z3-

Rotating about the point B, we get z_1 z_2z_3 z_2=a√(2)ae^iπ/4 = √(2)(1√(2) + i√(2))= (1+i) z_1 z_2 = (z_3 z_2)(1+i) or z_2 = z_2 (z_3 z_2)( ...

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Byju's Answer

Standard XII

Mathematics

Coordinates of a Point in Space

Az1, Bz2, Cz3...

Question

$A (z_{1}), B (z_{2}), C (z_{3})$ are the vertices of the triangle ABC (in anticlockwise order). If $∠ A B C = π / 4$ and $A B = \sqrt{2} (B C)$ .
Then prove that $z_{2} = z_{3} + i (z_{1} - z_{3}) .$

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Solution

Rotating about the point B, we get
$\frac{z_{1} - z_{2}}{z_{3} - z_{2}} = \frac{a \sqrt{2}}{a} e^{i π / 4} = \sqrt{2} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = (1 + i)$
$z_{1} - z_{2} = (z_{3} - z_{2}) (1 + i)$
or $z_{2} = z_{2} - (z_{3} - z_{2}) (1 + i)$
or $z_{2} (1 - (1 + i)) = z_{1} - z_{3} (1 + i)$
or $z_{2} = \frac{z_{1}}{- i} - \frac{z_{3}}{- i} (1 + i) = (i z_{1} - i z_{3} (1 + i)) = z_{3} + i (z_{1} - z_{3}) .$

Ans: 1

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A(z1),B(z2),C(z3) are the vertices of the triangle ABC (in anticlockwise order). If ∠ABC=π/4 and AB=√2(BC).Then prove that z2=z3+i(z1−z3).

Rotating about the point B, we getz1−z2z3−z2=a√2aeiπ/4=√2(1√2+i√2)=(1+i)z1−z2=(z3−z2)(1+i)or z2=z2−(z3−z2)(1+i)or z2(1−(1+i))=z1−z3(1+i)or z2=z1−i−z3−i(1+i)=(iz1−iz3(1+i))=z3+i(z1−z3). Ans: 1

$A (z_{1}), B (z_{2}), C (z_{3})$ are the vertices of the triangle ABC (in anticlockwise order). If $∠ A B C = π / 4$ and $A B = \sqrt{2} (B C)$ .
Then prove that $z_{2} = z_{3} + i (z_{1} - z_{3}) .$