Let \overset{\rightarrow}{a} = \hat{i} - 2 \hat{j} + \hat{k} \textrm{ } and \textrm{ } \overset{\rightarrow}{b} = \hat{i} - \hat{j} + \hat{k} be two vectors If \overset{\rightarrow}{c} is a vector such that \overset{\rightarrow}{b} \times \overset{\rightarrow}{c} = \overset{\rightarrow}{b} \times \overset{\rightarrow}{a} \textrm{ } and \textrm{ } \overset{\rightarrow}{c} . \overset{\rightarrow}{a} = 0 then\textrm{ } \overset{\rightarrow}{c} . \overset{\rightarrow}{b} is equal to

Engineering

Mathematics

Vector Triple Product

Question

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 $\vec{b} \times \vec{c} = \vec{b} \times \vec{a} and \vec{c} . \vec{a} = 0$ then $\vec{c} . \vec{b}$ is equal to

$- \frac{1}{2}$

$- \frac{3}{2}$

—1

$\frac{1}{2}$

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$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times (\vec{b} \times \vec{a})$

$- (\vec{a} . \vec{b}) \vec{c} = (\vec{a} . \vec{a}) \vec{b} - (\vec{a} . \vec{b}) \vec{a}$

$- 4 \vec{c} = 6 (\hat{i} - \hat{j} + \hat{k}) - 4 (\hat{i} - 2 \hat{j} + \hat{k})$

$- 4 \vec{c} = 2 \hat{i} + 2 \hat{j} + 2 \hat{k}$

$\vec{c} = - \frac{1}{2} (\hat{i} + \hat{j} + \hat{k})$

$\vec{b} . \vec{c} = - \frac{1}{2}$