Question

Angle that the a=2i+2j makes with y-axis is

Answer 1

Solution :

⏭ Given:

✏ $\sf \vec{A} = 2\hat{i} + 2\hat{j}$

⏭ To Find:

✏ Angle of given vector with y-axis

⏭ Formula:

✏ Formula of dot product of two vectors is given by

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{ \large{ \vec{A} { \tiny{ \bullet}} \vec{B} = | \vec{A}| | \vec{B}| \cos \Theta}}}}}} \: \star$

⏭ Calculation:

✏ Let, vector B lies on y-axis

$\therefore \sf \: \red{ \vec{B} = 1 \hat{j}} \\ \\ \leadsto \sf \: | \vec{A}| = \sqrt{ {2}^{2} + {2}^{2} } = 2 \sqrt{2} \: unit \\ \\ \leadsto\sf \: | \vec{B}| = \sqrt{ {1}^{2} } = 1 \: unit \\ \\ \implies \sf \: (2 \hat{i} + 2 \hat{j}) \: { \tiny{ \bullet}} \: 1 \hat{j} = (2 \sqrt{2} )(1) \cos \Theta \\ \\ \implies\sf \: (2 \times 1)( \hat{j} \times \hat{j}) = 2 \sqrt{2} \cos \Theta \\ \\ \implies \sf \: \cancel{2} = \cancel{2} \sqrt{2} \cos \Theta \\ \\ \implies \sf \: \cos \Theta = \dfrac{1}{ \sqrt{2} } \\ \\ \implies \sf \: \Theta = { \cos}^{ - 1} \dfrac{1}{ \sqrt{2} } \\ \\ \implies \: \underline{ \boxed{ \bold{ \sf{ \orange{ \large{ \Theta = 45 \degree}}}}}} \: \gray{ \bigstar}$

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• Second method :

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$\implies \sf \: \tan{ \Theta} = \dfrac{x \: component}{y \: component} \\ \\ \implies \sf \: \tan{ \Theta} = \frac{2}{2} = 1 \\ \\ \implies \: \boxed{ \purple{ \sf{ \Theta = 45 \degree}}}$