Question

Two vectors both equal in magnitude have their resultant equal to either. Find the angle between the two vectors.

Answer 1

Answer:

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Answer 2

Explanation:

Let {tex}\theta {/tex} be the angle between the two vectors A and B such that, |A| = |B|.

Let {tex}\theta {/tex} be the angle between the two vectors A and B such that, |A| = |B|.Therefore, {tex}R ^2 = A ^2 + B ^2 + 2AB cos\theta{/tex}

Let {tex}\theta {/tex} be the angle between the two vectors A and B such that, |A| = |B|.Therefore, {tex}R ^2 = A ^2 + B ^2 + 2AB cos\theta{/tex}{tex}=> A^2 = A ^2 + A ^2 + 2AAcos\theta{/tex}[{tex}\because {/tex} |A| = |B| = |R|]

Let {tex}\theta {/tex} be the angle between the two vectors A and B such that, |A| = |B|.Therefore, {tex}R ^2 = A ^2 + B ^2 + 2AB cos\theta{/tex}{tex}=> A^2 = A ^2 + A ^2 + 2AAcos\theta{/tex}[{tex}\because {/tex} |A| = |B| = |R|]{tex}=> A^ 2 = 2A^ 2 + 2A ^2 cos\theta {/tex}

Let {tex}\theta {/tex} be the angle between the two vectors A and B such that, |A| = |B|.Therefore, {tex}R ^2 = A ^2 + B ^2 + 2AB cos\theta{/tex}{tex}=> A^2 = A ^2 + A ^2 + 2AAcos\theta{/tex}[{tex}\because {/tex} |A| = |B| = |R|]{tex}=> A^ 2 = 2A^ 2 + 2A ^2 cos\theta {/tex}{tex}=> cos\theta ={ -1\over 2}\\ => \theta = 120^°{/tex}

Thnx.....