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Vector Algebra

Question
CBSEENMA12034024
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A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.

Solution
Let | OA | = 4 km and | AB | = 3 km then ∠OAB = 90° – 30° = 60°.

From B,  draw BM perpendicular OA.
Let  straight x space straight i with hat on top space plus space straight y space straight j with hat on top be the position vector of B.
therefore space space space space space space space space space space space space space space space space space space space space open vertical bar straight x close vertical bar space equals space open vertical bar OM close vertical bar space equals space open vertical bar OA close vertical bar space minus space open vertical bar MA close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 space minus space left parenthesis open vertical bar straight A space straight B close vertical bar space cos space 60 degree right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 minus 3 space open parentheses 1 half close parentheses space equals space 5 over 2 space therefore space space space space space space space space space space space space space space straight x space equals space minus 5 over 2
Also               straight y space equals space MB space equals space open vertical bar straight A thin space straight B close vertical bar space sin space 60 degree space equals space fraction numerator 3 space square root of 3 over denominator 2 end fraction.
therefore             OB with rightwards arrow on top space equals space straight x straight i with hat on top space plus space straight y space straight j with hat on top space equals space minus 5 over 2 straight i with hat on top space plus fraction numerator 3 square root of 3 over denominator 2 end fraction straight j with hat on top
Distance of B from initial point equals space open vertical bar OB close vertical bar space equals space open vertical bar OB with rightwards arrow on top close vertical bar
                               equals space open vertical bar negative 5 over 2 straight i with hat on top space plus space fraction numerator 3 square root of 3 over denominator 2 end fraction straight j with hat on top close vertical bar space equals space square root of open parentheses fraction numerator negative 5 over denominator 2 end fraction close parentheses squared plus space open parentheses fraction numerator 3 space square root of 3 over denominator 2 end fraction close parentheses squared end root equals space square root of 25 over 4 plus 27 over 4 end root space equals square root of fraction numerator 25 plus 27 over denominator 4 end fraction end root equals space square root of 52 over 4 end root space equals square root of 13 space km.

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