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Question
Find the angles which the vector \[\vec{a} = \hat{i} -\hat {j} + \sqrt{2} \hat{k}\] makes with the coordinate axes.
Solution
\[\text{Let } \theta_1\text{ be the angle between } \vec{a}\text { and } x - axis.\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( - 1 \right)^2 + \left( \sqrt{2} \right)^2} = \sqrt{4} = 2\]
\[ \vec{b} = \hat{i}\text { (Because } \hat{i} \text{ is the unit vector alongx-axis) }\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 1 + 0 + 0 = 1\]
\[\cos \theta_1 = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{1}{\left( 2 \right)\left( 1 \right)} = \frac{1}{2}\]
\[ \Rightarrow \theta_1 = \cos^{- 1} \left( \frac{1}{2} \right) = \frac{\pi}{3}\]
\[\]
\[\text{ Let } \theta_2 \text{ be the angle between } \vec{a} \text{ and } y - axis.\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( - 1 \right)^2 + \left( \sqrt{2} \right)^2} = \sqrt{4} = 2\]
\[ \vec{b} = \hat{j}............\text{ (Because } \hat{j}\text{ is the unit vector alongy-axis) }\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 0 - 1 + 0 = - 1\]
\[\cos \theta_2 = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{- 1}{\left( 2 \right)\left( 1 \right)} = \frac{- 1}{2}\]
\[ \Rightarrow \theta_2 = \cos^{- 1} \left( \frac{- 1}{2} \right) = \frac{2\pi}{3}\]
\[\]
\[\text{ Let } \theta_3\text{ be the angle between } \vec{a} \text{ and } z - axis.\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( - 1 \right)^2 + \left( \sqrt{2} \right)^2} = \sqrt{4} = 2\]
\[ \vec{b} = \hat{k}...............\text{ (Because } \hat{ k }\text { is the unit vector along z-axis) }\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 0 + 0 + \sqrt{2} = \sqrt{2}\]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{\sqrt{2}}{\left( 2 \right)\left( 1 \right)} = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}\]
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