Miscellaneous

Chapter 10 Class 12 Vector Algebra
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Misc 1 Write down a unit vector in XY-plane, making an angle of 30Β° with the positive direction of x-axis.Let the unit vector be π β π β = π₯π Μ + yπ Μ + zπ Μ Since the vector is in XY plane, there is no Z βcoordinate. π β = xπ Μ + yπ Μ + 0π Μ π β = ππ Μ + yπ Μ Since π β makes an angle 30Β° with the xβaxis Also, Unit vector in direction of x axis is π Μ & in y axis is π Μ Angle with X-axis Since π β makes an angle of 30Β° with x-axis So, angle between π β & π Μ is 30Β° We know that, π β . π β = |π β ||π β | cos ΞΈ, Putting π β = π β , π β = π Μ & ΞΈ = ΞΈ 30Β° π β .π Μ = |π β ||π Μ | cos 30Β° π β .π Μ = 1 Γ 1 Γ cos 30Β° π β . π Μ = cos 30Β° (π₯π Μ + yπ Μ + 0π Μ). π Μ = cos 30Β° (π₯π Μ + yπ Μ + 0π Μ). (1π Μ + 0π Μ + 0π Μ) = cos 30Β° π₯.1 + y.0 + 0.0 = cos 30Β° (As π β is unit vector, |π β | = 1 & π Μ is a unit vector, |π Μ | = 1) x = cos 30Β° x = βπ/π Angle with Y-axis π β makes an angle of (90Β° β 30Β°) i.e. 60Β° with y-axis So, angle between π β & π Μ is 60Β° We know that, π β . π β = |π β ||π β | cos ΞΈ, Putting π β = π β , π β = π Μ & ΞΈ = 60Β° π β .π Μ = |π β ||π Μ | cos 60Β° π β .π Μ = 1 Γ 1 Γ cos 60Β° π β .π Μ = cos 60Β° (π₯π Μ + yπ Μ + 0π Μ). π Μ = cos 60Β° (π₯π Μ + yπ Μ + 0π Μ). (0π Μ + 1π Μ + 0π Μ) = cos 60Β° π₯.0 + y.1 + 0.0 = cos 60Β° y = cos 60Β° y = π/π Thus, π β = xπ Μ + yπ Μ π β = βπ/π π Μ + π/π π Μ