If 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 toConsider all vectors are non-zero-

We have b×c=b×d⇒a× (b×c)=a× (b×d) ⇒ nbsp;(a·c)b (a·b)c=(a·d)b (a·b)d ⇒ nbsp;(a·b)d= (a·c)b+(a·b)c (∵a·d=0) ⇒d=c ( a·ca·b )b ...

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Byju's Answer

Standard XII

Physics

Vectors and Its Types

If the vector...

Question

If 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 $\to a \cdot \to d = 0 \cdot$ Then the vector $\to d$ is equal to
(Consider all vectors are non-zero.)

\to c - ⎛ ⎝ \frac{\to a \cdot \to c}{\to a \cdot \to b} ⎞ ⎠ \to b

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\to b - ⎛ ⎝ \frac{\to b \cdot \to c}{\to a \cdot \to b} ⎞ ⎠ \to c

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\to c + ⎛ ⎝ \frac{\to a \cdot \to c}{\to a \cdot \to b} ⎞ ⎠ \to b

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\to b + ⎛ ⎝ \frac{\to b \cdot \to c}{\to a \cdot \to b} ⎞ ⎠ \to c

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Solution

The correct option is A $\to c - ⎛ ⎝ \frac{\to a \cdot \to c}{\to a \cdot \to b} ⎞ ⎠ \to b$
We have $\to b \times \to c = \to b \times \to d \Rightarrow \to a \times (\to b \times \to c) = \to a \times (\to b \times \to d)$ $\Rightarrow (\to a \cdot \to c) \to b - (\to a \cdot \to b) \to c = (\to a \cdot \to d) \to b - (\to a \cdot \to b) \to d$ $\Rightarrow (\to a \cdot \to b) \to d = - (\to a \cdot \to c) \to b + (\to a \cdot \to b) \to c (∵ \to a \cdot \to d = 0) \Rightarrow \to d = \to c - ⎛ ⎝ \frac{\to a \cdot \to c}{\to a \cdot \to b} ⎞ ⎠ \to b$

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If 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 (Consider all vectors are non-zero.)

If 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 $\to a \cdot \to d = 0 \cdot$ Then the vector $\to d$ is equal to
(Consider all vectors are non-zero.)