Let z 1 = a 1 + ib 1 (a 1 , b 1 ) and z 2 = a 2 + ib 2 (a 2 , b 2 ); where i = -1 , be…
Mathematics · JEE Advanced · NTA Exams — Complex Numbers and Quadratic Equations
Let z1 = a1 + ib1 \(\equiv\)(a1, b1) and z2 = a2 + ib2 \(\equiv\) (a2, b2); where \(\mathrm{i}=\sqrt{-1},\) be two complex numbers.
If ∠POQ = θ, From Rotation theorem
\(\frac{z_{2}-0}{z_{1}-0}=\frac{\left|z_{2}\right|}{\left|z_{1}\right|} e^{i \theta} \Rightarrow \frac{z_{2} \bar{z}_{1}}{z_{1} \bar{z}}=\frac{\left|z_{2}\right|}{\left|z_{1}\right|} e^{i \theta}\)
⇒ \(\frac{\mathrm{z}_{2} \overline{\mathrm{z}}_{1}}{\left|\mathrm{z}_{1}\right|^{2}}=\frac{\left|\mathrm{z}_{2}\right|}{\left|\mathrm{z}_{1}\right|} \mathrm{e}^{\mathrm{i} \theta} \Rightarrow \mathrm{z}_{2} \overline{\mathrm{z}}_{1}=\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right| \mathrm{e}^{\mathrm{i} \theta}\)
⇒ \(z_{2} \bar{z}_{1}=\left|z_{1}\right|\left|z_{2}\right|(\cos \theta+i \sin \theta)\)
∴ \(\operatorname{Re}\left(\mathrm{z}_{2} \overline{\mathrm{z}}_{1}\right)=\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right| \cos \theta\) ... (i)
and \(\operatorname{Im}\left(\mathrm{z}_{2} \overline{\mathrm{z}}_{1}\right)=\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right| \sin \theta\) ... (ii)
The dot product of z1 and z2 is defined by z1 . z2 = |z1| |z2|
cos θ = Re \(\left(\mathrm{z}_{2} \overline{\mathrm{z}}_{1}\right)\)[from (i)] and cross product of z1 and z2 is defined z1 × z2 = |z1| |z2| sin θ = Im \(\left(\mathrm{z}_{2} \overline{\mathrm{z}}_{1}\right)\) [from Eq. (ii)]
If z1 = 3 + 4i and z2 = 4 + 3i, then the value of \(\sin \theta\left(\pi<\theta<\frac{3 \pi}{2}\right)\)is equal to
- \(-\frac{1}{7}\)
- \(-\frac{7}{25}\)
- \(-\frac{24}{25}\)
- \(-\frac{1}{25}\)
Answer
(B) - 7 25
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