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

Parametric Differentiation

If |z 1| =...

Question

If |z $_{1}$ | $=$ 1, |z $_{2}$ | $=$ 2, |z $_{3}$ | $=$ 3 and |9z $_{1}$ z $_{2}$ + 4z $_{1}$ z $_{3}$ + z $_{2}$ z $_{3}$ | $=$ 12, then the value of $| z_{1}$ + $z_{2}$ + $z_{3} |$ is

56

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

60

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

2

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

64

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

Open in App

Solution

The correct option is C $2$
Given $| z_{1} | = 1, | z_{2} | = 2, | z_{3} | = 3$ and $| 9 z_{1} z_{2} + 4 z_{1} z_{3} + z_{2} z_{3} | = 12$
$| 9 z_{1} z_{2} + 4 z_{1} z_{3} + z_{2} z_{3} | = 12$
$\Rightarrow | z_{1} | | z_{2} | | z_{3} | ∣ ∣ ∣ \frac{9}{z_{3}} + \frac{4}{z_{2}} + \frac{1}{z_{1}} ∣ ∣ ∣ = 12$
$\Rightarrow 6 ∣ ∣ ∣ ∣ \frac{9_{3}}{| z_{3} |^{2}} + \frac{4_{2}}{| z_{2} |^{2}} + \frac{_{1}}{| z_{1} |^{2}} ∣ ∣ ∣ ∣ = 12$
$\Rightarrow 6 | ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} + z_{2} + z_{3} | = 12$
$\Rightarrow 6 | z_{1} + z_{2} + z_{3} | = 12$
$∴ | z_{1} + z_{2} + z_{3} | = 2$

Suggest Corrections

Similar questions

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

1, 2, 3

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

| {9 z}_{1} z_{2} + z_{2} z_{3} + {4 z}_{3} z_{1} |

z_{1}, z_{2}

z_{3}

∣ ∣ ∣ ∣ \begin{matrix} z_{1} & z_{2} & 1 z_{2} & z_{3} & 1 z_{3} & z_{1} & 1 \end{matrix} ∣ ∣ ∣ ∣

Δ A B C

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

z_{1}^{'}, z_{2}^{'}, z_{3}^{'}

z_{3} = (1 - λ) z_{1} + λ z_{2}

z_{3}^{'} = (1 - μ) z_{1}^{'} + μ z_{2}^{'}

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

z, z_{1} a n d z_{2}

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

| z |

z

| z + \frac{2}{z} | = 2

1 + \sqrt{3}

| z_{1} + z_{2} | \leq | z_{1} | + | z_{2} |

Join BYJU'S Learning Program