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Question:
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Compute the magnitude of the following vectors:
$$
\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k}, \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}
$$
$$
\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k}, \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}
$$
Solution:
2070 Upvotes
Verified Answer
$$
\text { (i) } \begin{aligned}
\vec{a}=\hat{i}+\hat{j}+\hat{k}, &|\vec{a}|=\sqrt{1^2+1^2+1^2} \\
& {\left[\because|x \hat{i}+\hat{y}+z \hat{k}|=\sqrt{x^2+y^2+z^2}\right] }
\end{aligned}
$$
$$
\therefore \quad|\vec{a}|=\sqrt{3}
$$
(ii) $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
$$
\therefore|\overrightarrow{\mathrm{b}}|=\sqrt{2^2+(-7)^2+3^2}=\sqrt{62}
$$
(iii) $\overrightarrow{\mathrm{c}}=\frac{1}{\sqrt{3}} \hat{\mathrm{i}}+\frac{1}{\sqrt{3}} \hat{\mathrm{j}}-\frac{1}{\sqrt{3}} \hat{\mathrm{k}}$,
$$
|\overrightarrow{\mathrm{c}}|=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{-1}{\sqrt{3}}\right)^2}=1
$$
\text { (i) } \begin{aligned}
\vec{a}=\hat{i}+\hat{j}+\hat{k}, &|\vec{a}|=\sqrt{1^2+1^2+1^2} \\
& {\left[\because|x \hat{i}+\hat{y}+z \hat{k}|=\sqrt{x^2+y^2+z^2}\right] }
\end{aligned}
$$
$$
\therefore \quad|\vec{a}|=\sqrt{3}
$$
(ii) $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
$$
\therefore|\overrightarrow{\mathrm{b}}|=\sqrt{2^2+(-7)^2+3^2}=\sqrt{62}
$$
(iii) $\overrightarrow{\mathrm{c}}=\frac{1}{\sqrt{3}} \hat{\mathrm{i}}+\frac{1}{\sqrt{3}} \hat{\mathrm{j}}-\frac{1}{\sqrt{3}} \hat{\mathrm{k}}$,
$$
|\overrightarrow{\mathrm{c}}|=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{-1}{\sqrt{3}}\right)^2}=1
$$
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