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-

a . b 0 ( given ) #160; a . b =0 b × c = b × d ⇒ a ×( b × c #160; ) = a ×( b × d #160; ) #160; ⇒( a . c #160; ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

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 $\to a \cdot \to d = 0$ . Then the vector $\to d$ is equal to?

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Solution

$\to a . \to b \neq 0 (g i v e n) \to a . \to b = 0 \to b \times \to c = \to b \times \to d \Rightarrow \to a \times (\to b \times \to c) = \to a \times (\to b \times \to d) \Rightarrow (\to a . \to c) \to b - (\to a . \to b) \to c = (\to a . \to d) \to b - (\to a . \to b) \to d \Rightarrow (\to a . \to c) \to b - (\to a . \to b) \to c = 0 - (\to a . \to b) \to d \Rightarrow (\to a . \to b) \to d = - (\to a . \to c) \to b + (\to a . \to b) \to c \to d = - \frac{(\to a . \to c) \to b}{(\to a . \to b)} + \to c$

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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 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 $\to a \cdot \to d = 0$ . Then the vector $\to d$ is equal to?