Four vectors a- b- c and x satisfy the relation a-xb-c-x where b-a- 1- The value of x in terms of a- b and c is equal to

(a·x)b=c+x … (1) Taking dot product with a (a·x)(b·a)=(c·a)+(a·x) ⇒(a·x)(b·a 1)=c·a ⇒a·x=c·ab·a 1 … (2) From (1) and (2), ( c·ab·a 1 )b=c+x ⇒x= ( c·ab·a 1 ...

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Byju's Answer

Standard XII

Physics

Vectors and Its Types

Four vectors ...

Question

Four vectors $\to a, \to b, \to c$ and $\to x$ satisfy the relation $(\to a \cdot \to x) \to b = \to c + \to x$ where $\to b \cdot \to a \neq 1.$ The value of $\to x$ in terms of $\to a, \to b$ and $\to c$ is equal to

\frac{(\to a \cdot \to c) \to b - \to c (\to a \cdot \to b - 1)}{(\to a \cdot \to b - 1)}

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\frac{\to c}{\to a \cdot \to b - 1}

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\frac{2 (\to a \cdot \to c) \to b + \to c}{\to a . \to b - 1}

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\frac{2 (\to a \cdot \to c) \to c + \to c}{\to a \cdot \to b - 1}

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Solution

The correct option is A $\frac{(\to a \cdot \to c) \to b - \to c (\to a \cdot \to b - 1)}{(\to a \cdot \to b - 1)}$
$(\to a \cdot \to x) \to b = \to c + \to x \dots (1)$
Taking dot product with $\to a$
$(\to a \cdot \to x) (\to b \cdot \to a) = (\to c \cdot \to a) + (\to a \cdot \to x)$
$\Rightarrow (\to a \cdot \to x) (\to b \cdot \to a - 1) = \to c \cdot \to a$
$\Rightarrow \to a \cdot \to x = \frac{\to c \cdot \to a}{\to b \cdot \to a - 1} \dots (2)$

From $(1)$ and $(2),$
$⎛ ⎝ \frac{\to c \cdot \to a}{\to b \cdot \to a - 1} ⎞ ⎠ \to b = \to c + \to x \Rightarrow \to x = ⎛ ⎝ \frac{\to c \cdot \to a}{\to b \cdot \to a - 1} ⎞ ⎠ \to b - \to c = \frac{(\to a \cdot \to c) \to b - \to c (\to a \cdot \to b - 1)}{(\to a \cdot \to b - 1)}$

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Four vectors →a,→b,→c and →x satisfy the relation (→a⋅→x)→b=→c+→x where →b⋅→a≠1. The value of →x in terms of →a,→b and →c is equal to

Four vectors $\to a, \to b, \to c$ and $\to x$ satisfy the relation $(\to a \cdot \to x) \to b = \to c + \to x$ where $\to b \cdot \to a \neq 1.$ The value of $\to x$ in terms of $\to a, \to b$ and $\to c$ is equal to