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Equation of a Plane : Vector Form

Let A, B, C...

Question

Let $A, B, C$ be three collinear points which are such that $A B . A C = 1$ and the points are represented in the Argand plane by the complex numbers $0, z_{1}, z_{2}$ respectively. Then

A

z_{1}, z_{2} = 1

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B

z_{1} ._{2} = 1

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C

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

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D

none of these

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Solution

The correct options are
A $z_{1} ._{2} = 1$
B $| z_{1} | | z_{2} | = 1$
Since the points $0, z_{1}, z_{2}$ lie on a same line,
$a r g (z_{1}) = a r g (z_{2})$
Hence they must lie on a line of the form
$y = m x$
Now
$A B = z_{1}$
$A C = z_{2}$
$z_{1} . z_{2} = 1$
$z_{1} =_{2}$
However $z_{1}$ and $z_{2}$ have the same argument.
Hence this is only possible if $z_{1}$ and $z_{2}$ both are purely real.
Therefore
$z_{1} =_{1}$ and $z_{2} =_{2}$
Only $| z_{1} | = \frac{1}{| z_{2} |}$
Hence
$z_{1} ._{2} = z_{1} . z_{2} = 1$
And $| z_{1} | | z_{2} | = 1$

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Q

are represented by the complex numbers

z_{1}

z_{2}

respectively, such that

∣ ∣ ∣ \frac{1}{z_{2}} + \frac{1}{z_{1}} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{1}{z_{2}} - \frac{1}{z_{1}} ∣ ∣ ∣

Q. The triangle whose vertices are the points represented by the complex numbers

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

on the Argand diagram is equilateral. Then show that

\frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} + \frac{1}{z_{1} - z_{2}} = 0

z_{1}

z_{2}

be two non zero complex numbers such that

\frac{z_{1}}{z_{2}} + \frac{z_{2}}{z_{1}} = 1

, then the origin and the points represented by

z_{1}

z_{2}

z_{1}, z_{2}

z_{1} + z_{2}

represents three points

A, B

C

in Argand plane. If

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

, then prove that the area of triangle

A B C

\frac{\sqrt{3}}{4} | z_{1} |^{2}

Q. If the points

A

B

are represented by the non-zero complex number

z_{1}

z_{2}

on the argand plane such that

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

O

is the origin, then,

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