Let a⇀=2i^-3j^+4k^ and b⇀=7i^+j^-6k^. If r⇀×a⇀=r⇀×b⇀, r⇀.i^+2j^+k^=-3, then r⇀.2i^-3j^+k^ is equal to

r¯×a¯=r¯×b¯⇒r¯×(a¯-b¯)=0 ⇒r¯=λ(a¯-b¯)=λ(-5i-4j+10k)r¯.(i+2j+k)=-3⇒λ(-5-8+10)=-3⇒λ=1r¯.(2i-3j+k)=1(-10+12+10)=12

Let a⇀=2i^-3j^+4k^ and b⇀=7i^+j^-6k^. If r⇀×a⇀=r⇀×b⇀, r⇀.i^+2j^+k^=-3, then r⇀.2i^-3j^+k^ is equal to

r¯×a¯=r¯×b¯⇒r¯×(a¯-b¯)=0 ⇒r¯=λ(a¯-b¯)=λ(-5i-4j+10k)r¯.(i+2j+k)=-3⇒λ(-5-8+10)=-3⇒λ=1r¯.(2i-3j+k)=1(-10+12+10)=12

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$Let \overset{⇀}{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} and \overset{⇀}{b} = 7 \hat{i} + \hat{j} - 6 \hat{k} . If \overset{⇀}{r} \times \overset{⇀}{a} = \overset{⇀}{r} \times \overset{⇀}{b}, \overset{⇀}{r} . (\hat{i} + 2 \hat{j} + \hat{k}) = - 3, then \overset{⇀}{r} . (2 \hat{i} - 3 \hat{j} + \hat{k}) is equal to$

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Detailed Solution

$\begin{array}{rcl} \bar{r} \times \bar{a} & = & \bar{r} \times \bar{b} \Rightarrow \bar{r} \times (\bar{a} - \bar{b}) = 0 \\ \Rightarrow & \bar{r} = λ (\bar{a} - \bar{b}) = λ (- 5 i - 4 j + 10 k) \\ \bar{r} . (i + 2 j + k) & = & - 3 \Rightarrow λ (- 5 - 8 + 10) = - 3 \Rightarrow λ = 1 \\ \bar{r} . (2 i - 3 j + k) & = & 1 (- 10 + 12 + 10) = 12 \end{array}$