Question

A= 2i+3j
B= -i+2j
Find the angle between A and B vectors

Answer 1

Topic :- Vectors

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Given;

$\footnotesize\bullet\: \underline{\sf \overrightarrow{A} = 2 \hat{i} + 3\hat{j}}$

$\footnotesize\bullet\: \underline{\sf \overrightarrow{B} = - \hat{i} + 2\hat{j}}$

We need to find the angle between this two vectors i.e vector A and vector B.

$\longrightarrow\:\sf \cos (\theta) =\dfrac{\overrightarrow{A}.\overrightarrow{B}}{AB}$

$\longrightarrow\:\sf \cos (\theta) =\dfrac{(2 \hat{i} + 3\hat{j}).( - \hat{i} + 2\hat{j})}{ \sqrt{ {2}^{2} + {3}^{2} }. \sqrt{ {1}^{2} + {2}^{2} } }$

$\longrightarrow\:\sf \cos (\theta) =\dfrac{ - 2 + 6}{ \sqrt{ 4 + 9 }. \sqrt{ 1 + 4 } }$

$\longrightarrow\:\sf \cos (\theta) =\dfrac{ 4}{ \sqrt{ 4 + 9 }. \sqrt{ 1 + 4 } }$

$\longrightarrow\:\sf \cos (\theta) =\dfrac{ 4}{ \sqrt{ 13 }. \sqrt{ 5 } }$

$\longrightarrow\:\sf \cos (\theta) =\dfrac{ 4}{ \sqrt{ 65 } }$

$\longrightarrow\: \underline{ \underline{\sf (\theta) =\cos^{-1}\Bigg(\dfrac{ 4}{ \sqrt{ 65}} \Bigg)}}$

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Answer 2

Answer:

Angle, ∅ = cos⁻¹ 4/√65

Explanation:

Given;-

      A = 2i + 3j

      B = -i + 2j

Now. We know from dot product of vectors;-

         A.B = AB cos∅

So,  cos∅ = AB/ A.B

      cos∅ = ( 2i + 3j )( -i + 2j )/ √(2)² + (3)² . √1² + (2)²

      cos∅ = - 2 + 6/ √13 . √5

      cos∅ = 4 / √65

            ∅ = cos⁻¹ 4/√65

Therefore, the angle between A and B is ∅ = cos⁻¹ 4/√65

Hope it helps ;-))