The vectors a- and b- are not perpendicular and c- and d- are two vectors satisfying - b-c- - b-d- and a-d- - 0- Then the vector d- is equal to

The correct option is B c⃗ ( a⃗·c⃗a⃗ . b⃗) b⃗ a⃗·b⃗≠ 0 (given) a⃗·d⃗=0 Now, b⃗×c⃗=b⃗×d⃗⇒a⃗× (b⃗×c⃗)=a⃗× (b⃗×d⃗) ⇒ (a⃗·c⃗)b⃗ (a⃗·b⃗)·c ...

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L'Hospital Rule to Remove Indeterminate Form

The vectors ...

Question

The vectors $\to a$ and $\to b$ are not perpendicular and $\to c$ and $\to d$ are two vectors satisfying : $\to b \times \to c = \to b \times \to d$ and $\to a \times \to d = 0.$ Then the vector $\to d$ is equal to

\to b + (\frac{\to b . \to c}{\to a . \to b}) \to c

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\to c - (\frac{\to a \cdot \to c}{\to a . \to b}) \to b

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\to b - (\frac{\to b . \to c}{\to a . \to b}) \to c

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\to c + (\frac{\to a \cdot \to c}{\to a . \to b}) \to b

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Similar questions

\to a \times \to b = \to c \times \to d

\to a \times \to c = \to b \times \to d

\to a - \to d

\to b - \to c

\to a, \to b, \to c, \to d

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

\to b, \to c, \to d

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

\to b, \to c, \to d

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

\to a

\to b

\to c

\to d

\to b \times \to c = \to b \times \to d

\to a \cdot \to d = 0

\to d

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