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The magnitude of the scalar $\mathrm{p}$ for which the vector
$\mathrm{p}(-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})$ is of unit length is :
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$\mathrm{p}(-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})$ is of unit length is :
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{182}}$
Let $\overrightarrow{\mathrm{a}}=\mathrm{p}(-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})$
$=(-3 p) \hat{i}+(-2 p) \hat{j}+(13 p) \hat{k}$
It is given that a is of unit length
$\therefore|\overrightarrow{\mathrm{a}}|=1 \Rightarrow|\overrightarrow{\mathrm{a}}|^{2}=1$
$\Rightarrow \quad(-3 \mathrm{p}))(-3 \mathrm{p})+(-2 \mathrm{p})(-2 \mathrm{p})+(13 \mathrm{p})(13 \mathrm{p})=1$
$\quad 9 \mathrm{p}^{2}+4 \mathrm{p}^{2}+169 \mathrm{p}^{2}=1$
$\Rightarrow \quad \mathrm{p}^{2}=\frac{1}{182} \Rightarrow \mathrm{p}=\frac{1}{\sqrt{182}}$
$=(-3 p) \hat{i}+(-2 p) \hat{j}+(13 p) \hat{k}$
It is given that a is of unit length
$\therefore|\overrightarrow{\mathrm{a}}|=1 \Rightarrow|\overrightarrow{\mathrm{a}}|^{2}=1$
$\Rightarrow \quad(-3 \mathrm{p}))(-3 \mathrm{p})+(-2 \mathrm{p})(-2 \mathrm{p})+(13 \mathrm{p})(13 \mathrm{p})=1$
$\quad 9 \mathrm{p}^{2}+4 \mathrm{p}^{2}+169 \mathrm{p}^{2}=1$
$\Rightarrow \quad \mathrm{p}^{2}=\frac{1}{182} \Rightarrow \mathrm{p}=\frac{1}{\sqrt{182}}$
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