Misc 3 - A girl walks 4 km west, then 3 km 30 east of north

Misc 3 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 3 - Chapter 10 Class 12 Vector Algebra - Part 3

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Transcript

Misc 3 A girl walks 4 km towards west, then she walks 3 km in a direction 30Β° east of north and stops. Determine the girl’s displacement from her initial point of departure.Let initial point of departure be O(0, 0) & OA = 4km, AB = 3km ∠ XAB = 30Β° Here, Displacement will be vector (𝑂𝐡) βƒ— So, we need to find coordinates of B Rough Representation of 30Β° east of north Now, ∠ BAO = 90Β° – ∠ XAB = 90Β° – 30Β° = 60Β° Let us draw BD βŠ₯ OA In right triangle ADB sin 60Β° = 𝐡𝐷/𝐴𝐡 √3/2 = 𝐡𝐷/3 (3√3)/2 = BD BD = (πŸ‘βˆšπŸ‘)/𝟐In right triangle ADB cos 60Β° = 𝐴𝐷/𝐴𝐡 1/2 = 𝐴𝐷/3 3/2 = AD AD = πŸ‘/𝟐 Now, OD = OA – AD OD = 4 – πŸ‘/𝟐 = (πŸ’(𝟐) βˆ’ πŸ‘)/𝟐 = πŸ“/𝟐 So, x-coordinate of point B = βˆ’OD = (βˆ’πŸ“)/𝟐 Thus, coordinates of point B = ((βˆ’πŸ“)/𝟐, (πŸ‘βˆšπŸ‘)/𝟐) So, Displacement = (βˆ’πŸ“)/𝟐 π’Š Μ‚ + (πŸ‘βˆšπŸ‘)/𝟐 𝒋 Μ‚ y-coordinate of point B = BD = (πŸ‘βˆšπŸ‘)/𝟐

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.