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Cross Product of Two Vectors

The scalars l...

Question

The scalars l and m such that $l \to a + m \to b = \to c$ where $\to a, \to b$ and $\to c$ are given vectors, are equal to

A

$l = \frac{(\to c \times \to b) . (\to a \times \to b)}{| \to a \times \to b |^{2}}$ ; $m = \frac{(\to c \times \to a) . (\to b \times \to a)}{| \to b \times \to a |^{2}}$

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B

$l = \frac{(\to c \times \to b) . (\to a \times \to b)}{| \to a \times \to b |}$ ; $m = \frac{(\to c \times \to a) . (\to b \times \to a)}{| \to b \times \to a |}$

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C

$l = \frac{| \to c \times \to b |^{2}}{| \to a \times \to b |^{2}}$ ; $m = \frac{| \to c \times \to a |^{2}}{| \to b \times \to a |^{2}}$

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D

$l = \frac{| \to c \times \to a |^{2}}{| \to b \times \to a |^{2}}$ ; $m = \frac{| \to c \times \to b |^{2}}{| \to a \times \to b |^{2}}$

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Solution

The correct option is A
$l = \frac{(\to c \times \to b) . (\to a \times \to b)}{| \to a \times \to b |^{2}}$ ; $m = \frac{(\to c \times \to a) . (\to b \times \to a)}{| \to b \times \to a |^{2}}$

$l \to a + m \to b = \to c$
$(l \to a + m \to b) \times \to b = \to c \times \to b \Rightarrow l \to a \times \to b = \to c \times \to b$
Take dot product with $\to a \times \to b$
$\Rightarrow l | \to a \times \to b |^{2} = (\to c \times \to b) . (\to a \times \to b) \Rightarrow l = \frac{(\to c \times \to b) . (\to a \times \to b)}{| \to a \times \to b |^{2}}$
Similarly, $m = \frac{(\to c \times \to a) . (\to b \times \to a)}{| \to b \times \to a |^{2}}$

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Similar questions

The scalars l and m such that $l \to a + m \to b = \to c$ where $\to a, \to b$ and $\to c$ are given vectors, are equal to

\to A = \to b \times \to c, \to B = \to c \times \to a, \to C = \to a \times \to b

, then the vectors

\to A \times (\to B \times \to C)

\to B \times (\to C \times \to A)

\to C \times (\to A \times \to B)

[\to a + 2 \to b - \to c, \to a - \to b, \to a - \to b - \to c] =

| \to a | = 1, ∣ ∣ \to b ∣ ∣ = \sqrt{2}

| \to c | = \sqrt{3}

\to a ⊥ (\to b + \to c)

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∣ ∣ \to a + \to b + \to c ∣ ∣

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