For any vector r→,(r→⋅i^)2+(r→⋅j^)2+(r→⋅k^)2 is equal to

Let r→=xi^+yj^+zk^.Then,r→⋅i^=x,r→⋅j^=y and r→⋅k^=z∴ (r→⋅i^)2+(r→⋅i^)2+(r→⋅k^)2=x2+y2+z2=|r→|2

For any vector r→,(r→⋅i^)2+(r→⋅j^)2+(r→⋅k^)2 is equal to

Let r→=xi^+yj^+zk^.Then,r→⋅i^=x,r→⋅j^=y and r→⋅k^=z∴ (r→⋅i^)2+(r→⋅i^)2+(r→⋅k^)2=x2+y2+z2=|r→|2

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For any vector $\vec{r}, (\vec{r} \cdot \hat{i})^{2} + (\vec{r} \cdot \hat{j})^{2} + (\vec{r} \cdot \hat{k})^{2}$ is equal to

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$| \vec{r} |^{2}$

$| \vec{r} |$

$\vec{r}$

answer is D.

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Let $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k} .$ Then,

$\vec{r} \cdot \hat{i} = x, \vec{r} \cdot \hat{j} = y$ and $\vec{r} \cdot \hat{k} = z$

$∴ (\vec{r} \cdot \hat{i})^{2} + (\vec{r} \cdot \hat{i})^{2} + (\vec{r} \cdot \hat{k})^{2} = x^{2} + y^{2} + z^{2} = | \vec{r} |^{2}$

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For any vector r→,(r→⋅i^)2+(r→⋅j^)2+(r→⋅k^)2 is equal to

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For any vector $\vec{r}, (\vec{r} \cdot \hat{i})^{2} + (\vec{r} \cdot \hat{j})^{2} + (\vec{r} \cdot \hat{k})^{2}$ is equal to