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प्रश्न
If \[\overrightarrow{a} = \hat{i} + \hat{j} , \vec{b} = \hat{j} + \hat{k} \text{ and }\vec{c} = \hat{k} + \hat{i} ,\] write unit vectors parallel to \[\overrightarrow{a} + \overrightarrow{b} - 2 \overrightarrow{c} .\]
योग
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उत्तर
Given: \[\vec{a} = \hat{i} + \hat{j} , \vec{b} = \hat{j} + \hat{k} , \overrightarrow{c} = \hat{k} + \hat{i}\]
Now,
\[\overrightarrow{a} + \overrightarrow{b} - 2 \overrightarrow{c} = \hat{i} + \hat{j} + \hat{j} + \hat{k} - 2 \hat{k} - 2 \hat{i}\]
\[= - \hat{i} + 2 \hat{j} - \hat{k}\]
Unit vector parallel to \[\overrightarrow{a} + \overrightarrow{b} - 2 \overrightarrow{c} = \frac{- \hat{i} + 2 \hat{j} - \hat{k}}{\sqrt{\left( - 1 \right)^2 + 2^2 + \left( - 1 \right)^2}}\]
\[= \frac{- \hat{i} + 2 \hat{j} - \hat{k}}{\sqrt{6}}\]
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