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Let \vec{a}=\hat{i}-2\hat{j}+\hat{k} and \vec{b}=\hat{i}-\hat{j}+\hat{k} be two vectors. If \vec{c} is a vector such that \vec{b}\times \vec{c}=\vec{b}\times \vec{a} and \vec{c}\cdot \vec{a}=0, then \vec{c}\cdot \vec{b} is equal to :
Option: 1 \frac{1}{2}
Option: 2-\frac{3}{2}
Option: 3 -\frac{1}{2}
Option: 4 -1

Answers (1)

best_answer

 

 

Vector Triple Product -

For three  vectors \vec {\mathbf a},\;\vec{\mathbf b} and \vec {\mathbf c} vector triple product is defined as \vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right ).
\vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )=\left ( \vec {\mathbf a}\cdot \vec {\mathbf c} \right )\cdot \vec {\mathbf b}\;-\;\left ( \vec {\mathbf a}\cdot \vec {\mathbf b} \right )\cdot \vec {\mathbf c}

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\\\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}) \\ {\quad-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}=(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}} \\ {\quad\quad\quad\quad-4 \overrightarrow{\mathrm{c}}=6(\hat{\mathrm{i}}-\hat{j}+\hat{\mathrm{k}})-4(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})} \\ {\quad\quad\quad\quad-4 \overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}} \\ {\quad\quad\quad\quad\quad\overrightarrow{\mathrm{c}}=-\frac{1}{2}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})} \\ {\quad\quad\quad\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-\frac{1}{2}}

Correct Option (3)

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