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Let $\alpha$ be the angle which the vector $\overrightarrow{\mathrm{V}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ makes
with the z-axis, Then, what is the value of $\sin \alpha$ ?
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with the z-axis, Then, what is the value of $\sin \alpha$ ?
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Verified Answer
The correct answer is:
$\frac{\sqrt{5}}{3} \quad$
The given vector is $\overrightarrow{\mathrm{V}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
and for z-axis $\mathrm{x}=0$ and $\mathrm{y}=0$, so the vector equation is $\overrightarrow{\mathrm{A}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\cos \alpha=\frac{\overrightarrow{\mathrm{V}} \cdot \overrightarrow{\mathrm{A}}}{|\overrightarrow{\mathrm{V}}| \cdot|\overrightarrow{\mathrm{A}}|}$
$\cos \alpha=\frac{2.0+(-1) \cdot(0)+2 \cdot(1)}{\sqrt{4+1+4} \sqrt{0+0+1}}=\frac{2}{3}$
Hence, $\sin \alpha=\sqrt{1-\cos ^{2} \alpha}$
$=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}$
and for z-axis $\mathrm{x}=0$ and $\mathrm{y}=0$, so the vector equation is $\overrightarrow{\mathrm{A}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\cos \alpha=\frac{\overrightarrow{\mathrm{V}} \cdot \overrightarrow{\mathrm{A}}}{|\overrightarrow{\mathrm{V}}| \cdot|\overrightarrow{\mathrm{A}}|}$
$\cos \alpha=\frac{2.0+(-1) \cdot(0)+2 \cdot(1)}{\sqrt{4+1+4} \sqrt{0+0+1}}=\frac{2}{3}$
Hence, $\sin \alpha=\sqrt{1-\cos ^{2} \alpha}$
$=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}$
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