If l 1 2 - m 1 2 - n 1 2 -1 and l 1 l 2 - m 1 m 2 - n 1 n 2 -0 and - - l 1 m 1 n 1 l 2 m 2 n 2 l 3 m 3 n 3 then

The correct option is D | Δ| =1 Consider, Δ nbsp;^ 2 = l_ 1 amp; m_ 1 amp; n_ 1 l_ 2 amp; m_ 2 amp; n_ 2 l_ 2 amp; m_ 3 ...

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Evaluation of a Determinant

If l 1 2 + ...

Question

If $l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1$ and $l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$ and $Δ = ∣ ∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{3} & m_{3} & n_{3} \end{matrix} ∣ ∣ ∣ ∣$ then

| Δ | = 3

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| Δ | = 2

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| Δ | = 1

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Δ = 0

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Solution

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Similar questions

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos θ / 2 & 1 & 1 1 & cos θ / 2 & - cos θ / 2 - cos θ / 2 & 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ

m_{1}

Δ

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[m_{1}, m_{2}]

l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}; l_{3}, m_{3}, n_{3}

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

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

k

Δ

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} c o s \frac{θ}{2} & 1 & 1 1 & c o s \frac{θ}{2} & - c o s \frac{θ}{2} - c o s \frac{θ}{2} & 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

Δ

m_{1}

Δ

m_{2}

m_{1}

m_{2}

[lim x \to 0 \frac{sin x}{x}] = 0

x \in (- δ, δ)

δ

δ \to 0, tan x > x .

Δ

\sim

Δ

\frac{A B}{P Q} = \frac{1}{3}

\frac{a r Δ A B C}{a r Δ P Q R}

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