Given than q2 - pr 0 the value of p q px - qy q r qx - ry px - qy qx - ry 0 is

The correct option is C negative nbsp; p amp; q amp; px + qy q amp; r amp; qx + ry px + qy nbsp; amp; qx + ry nbsp; amp; 0 ...

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Definition of Vector

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Question

Given than $q^{2} - p r < 0, p > 0$ the value of $∣ ∣ ∣ ∣ \begin{matrix} p & q & p x + q y q & r & q x + r y p x + q y & q x + r y & 0 \end{matrix} ∣ ∣ ∣ ∣$ is

zero

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positive

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negative

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q^{2} + p r

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Solution

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Similar questions

q^{2} - p r < 0, p > 0

∣ ∣ ∣ ∣ \begin{matrix} p & q & p x + q y q & r & q x + r y p x + q y & q x + r y & 0 \end{matrix} ∣ ∣ ∣ ∣

p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0, p \neq 0

Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent, if

x^{3} + 3 p x^{2} + 3 q x + r

x^{2} + 2 p x + q

4 (p^{2} - q) (q^{2} - p r) - (p q - r)^{2} = 0

p^{2} - q = 0, q^{2} - p r = 0

q tan A = p

\frac{p sin A - q cos A}{q sin A + q cos A}

\frac{p^{2} - q^{2}}{p^{2} + q^{2}}

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