If the lines r-a- b-c and r-b- c-a intersects - - are scalars then

The correct option is D a·c=b·c For point of intersection r {a+α ( b×c ) } =r {b+β ( c×a ) } ⇒ #160; #160; a·c+α ( b×c )c=b·c+β ( ...

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Dot Product

If the lines ...

Question

If the lines $\to r = \to a + α (\to b \times \to c)$ and $\to r = \to b + β (\to c \times \to a)$ intersects ( $α$ , $β$ are scalars) then

\to a \cdot \to c = 0

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\to a \cdot \to c = \to b \cdot \to c

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\to b \cdot \to c = 0

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None of these

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\to d

\to d = λ \to b \times \to c + μ \to c \times \to a + ν \to a \times \to b

∣ ∣ ∣ (\to d \cdot \to a) (\to b \times \to c) + (\to d \cdot \to b) (\to c \times \to a) + (\to d \cdot \to c) (\to a \times \to b) ∣ ∣ ∣

[\to a \times \to b \to b \times \to c \to c \times \to a] = λ [\to a \to b \to c]^{2}

λ

sin (120^{0} - α) = sin (120^{0} - β), 0 < α, β < π

α

β

α

β

\to β . (\to α \times \to β) = \to 0

α

β

x^{2} + p x + q = 0

{(α - β)}^{2}

{(α + β)}^{2}

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