Chapter 10 Class 12 Vector Algebra
Concept wise

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Misc 4 If π β = π β + π β , then is it true that |π β|=|π β| + |π β| Justify your answer.Given, π β = π β + π β Let π β = 1π Μ + 2π Μ + 3π Μ & π β = 2π Μ β 1π Μ β 2π Μ Thus, π β = (π β + π β) = (1 + 2) π Μ + (2 β 1) π Μ + (3 β 2) π Μ = 3π Μ + 1π Μ + 1π Μ Given, π β = π β + π β Let π β = 1π Μ + 2π Μ + 3π Μ & π β = 2π Μ β 1π Μ β 2π Μ Thus, π β = (π β + π β) = (1 + 2) π Μ + (2 β 1) π Μ + (3 β 2) π Μ = 3π Μ + 1π Μ + 1π Μ Given, π β = π β + π β Let π β = 1π Μ + 2π Μ + 3π Μ & π β = 2π Μ β 1π Μ β 2π Μ Thus, π β = (π β + π β) = (1 + 2) π Μ + (2 β 1) π Μ + (3 β 2) π Μ = 3π Μ + 1π Μ + 1π Μ Finding |π β |, |π β | , |π β | Magnitude of π β = β(32+1^2+1^2 ) |π β | = β(9+1+1) = βππ Magnitude of π β = β(12+22+32) |π β | = β(1+4+9) = βππ Magnitude of π β = β(22+(β1)2+(β2)2) |π β | = β(4+1+4) = β9 = 3 Now, |π β | + |π β | = β14 + 3 β  β11 β  |π β | So, |π β |β  |π β | + |π β | Hence, the given statement is False.