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Algebra of Complex Numbers

If l1, m1, ...

Question

If $l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}; l_{3}, m_{3}, n_{3}$ are real quantities satisfying the six relations
$l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = l_{2}^{2} + m_{2}^{2} + n_{2}^{2} = l_{3}^{2} + m_{3}^{2} + n_{3}^{2} = 1$
$l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} = l_{3} l_{1} + m_{3} m_{1} + n_{3} n_{1} = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$
and $∣ ∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{3} & m_{3} & n_{3} \end{matrix} ∣ ∣ ∣ ∣ = \pm k$ . Find the value of $k$ .

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Solution

Let $Δ = ∣ ∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{3} & m_{3} & n_{3} \end{matrix} ∣ ∣ ∣ ∣$
$Δ^{2} = ∣ ∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{3} & m_{3} & n_{3} \end{matrix} ∣ ∣ ∣ ∣ \times ∣ ∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{3} & m_{3} & n_{3} \end{matrix} ∣ ∣ ∣ ∣$
Multiplying row by row
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} l_{1}^{2} + m_{1}^{2} + n_{1}^{2} & l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} & l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3} l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} & l_{1}^{2} + m_{2}^{2} + n_{2}^{2} & l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3} & l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} & l_{3}^{2} + m_{3}^{2} + n_{3}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣$ using given conditions.
$\Rightarrow Δ^{2} = 1$

$∴ Δ = \pm 1$

$∴ k = 1$

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Similar questions

l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1

l_{2}^{2} + m_{2}^{2} + n_{2}^{2} = 1

l_{3}^{2} + m_{3}^{2} + n_{3}^{2} = 1

l_{3} l_{1} + m_{3} m_{1} + n_{2} n_{1} = 0

∣ ∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{2} & m_{3} & n_{3} \end{matrix} ∣ ∣ ∣ ∣

Q. Show the matrix

⎡ ⎢ ⎣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{3} & m_{3} & n_{3} \end{matrix} ⎤ ⎥ ⎦

is orthogonal if

l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = \sum l_{1}^{2} = 1 = \sum l_{2}^{2} = \sum l_{3}^{2}

l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = \sum l_{1} l_{2} = 0 = \sum l_{2} l_{3} = \sum l_{3} l_{1}

l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1

l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0

Δ = ∣ ∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{3} & m_{3} & n_{3} \end{matrix} ∣ ∣ ∣ ∣

(l_{1}, m_{1}, n_{1})

(l_{2}, m_{2}, n_{2})

are d.c's two lines. Then
The lines are perpendicular if

l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0

l_{1}, m_{1}, n_{1}

l_{2}, m_{2}, n_{2}

are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines, will be

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