If a¯, b¯ are not perpendicular to each other and r¯×b¯=c¯×b¯, r¯.a¯=0, then r¯=

r×b=c×b r·a=0a×(r×b)=a×(c×b)(a·b) r-0=(a·b) c-(a·c) br=c-a·ca·b b

If a¯, b¯ are not perpendicular to each other and r¯×b¯=c¯×b¯, r¯.a¯=0, then r¯=

r×b=c×b r·a=0a×(r×b)=a×(c×b)(a·b) r-0=(a·b) c-(a·c) br=c-a·ca·b b

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If $\bar{a}, \bar{b}$ are not perpendicular to each other and $\bar{r} \times \bar{b} = \bar{c} \times \bar{b}, \bar{r} . \bar{a} = 0, then \bar{r} =$

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$\bar{c} - \frac{\bar{c} . \bar{a}}{\bar{a} . \bar{b}}$

$\bar{c} + \frac{\bar{a} . \bar{b}}{\bar{c} . \bar{a}}$

$\bar{c} - \frac{\bar{c} . \bar{a}}{\bar{a} . \bar{b}} \bar{b}$

$\bar{c} - \frac{\bar{a} . \bar{b}}{\bar{c} . \bar{a}} \bar{b}$

answer is C.

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$\begin{array}{rcl} r \times b & = & c \times b r \cdot a = 0 \\ a \times (r \times b) & = & a \times (c \times b) \\ (a \cdot b) r - 0 & = & (a \cdot b) c - (a \cdot c) b \\ r & = & c - (\frac{a \cdot c}{a \cdot b}) b \end{array}$

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If a¯, b¯ are not perpendicular to each other and r¯×b¯=c¯×b¯, r¯.a¯=0, then r¯=

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If $\bar{a}, \bar{b}$ are not perpendicular to each other and $\bar{r} \times \bar{b} = \bar{c} \times \bar{b}, \bar{r} . \bar{a} = 0, then \bar{r} =$