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If a = \hat { i } + \hat { j } + 2 \hat { k } \text { and } b = 2 \hat { i } + \hat { j } - 2 \hat { k } \\  find the unit vector in the direction of 2 \vec { a } - \vec { b }

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We need to find the unit vector in the direction of 2 \vec { a } - \vec { b }

First, let us calculate .2 \vec { a } - \vec { b }

As we have,

\begin{array}{l} \overrightarrow{\mathrm{a}}=\hat{1}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\ldots}(\mathrm{a}) \\ \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{\imath}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}_{\ldots(\mathrm{b})} \end{array}

Then multiply equation (a) by 2 on both sides,

2 \vec { a } = 2 ( \hat { \imath } + \hat { \jmath } + 2 \hat { k } )

We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.

\Rightarrow 2 \vec { a } = 2 \hat { \imath } + 2 \hat { \jmath } + 4 \hat { k } $ \ldots $ (c)\\

Subtract (b) from (c). We get

\Rightarrow 2 \vec { a } - \vec { b } = 2 \hat { l } - 2 \hat { l } + 2 \hat { j } - \hat { j } + 4 \hat { k } + 2 \hat { k } \\

\Rightarrow 2 \vec{a}-\vec{b}=\hat{\jmath}+6 \hat{k}

We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.

For finding unit vector, we have the formula:

\\ 2 \hat{a}-\hat{b}=\frac{2 \vec{a}-\vec{b}}{|2 \vec{a}-\vec{b}|}$ \\Now we know the value of $2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}},\\$ so we just need to substitute in the above equation.\\ $\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{|\hat{\jmath}+6 \hat{k}|}$\\ Here, $|\hat{\jmath}+6 \hat{\mathrm{k}}|=\sqrt{1^{2}+6^{2}}$ \\$\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{1^{2}+6^{2}}}

\begin{aligned} &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{\mathrm{k}}}{\sqrt{1+36}}\\ &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}}\\ &\text { Thus, unit vector in the direction of }\\ &2 \vec{a}-\vec{b}_{i s} \frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}} \end{aligned}

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