Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Example 14 Find angle βΞΈβ between the vectors π β = π Μ + π Μ β π Μ and π β = π Μ β π Μ + π Μ. Given π β = π Μ + π Μ β π Μ π β = π Μ β π Μ + π Μ We know that π β . π β = "|" π β"|" "|" π β"|" cos ΞΈ where ΞΈ is the angle between π β and π β Finding |π β |, |π β | and π β . π β Magnitude of π β = β(12+1^2+(β1)2) |π β | = β(1+1+1) = βπ Magnitude of π β = β(12+(β1)2+12) |π β | = β(1+1+1) = βπ Finding π β . π β π β . π β = (1π Μ + 1π Μ β 1π Μ). (1π Μ β 1π Μ + 1π Μ) = 1.1 + 1.(β1) + (β1)1 = 1 β 1 β 1 = β1 Now, π β . π β = "|" π β"|" "|" π β"|" cos ΞΈ Putting values β1 = β3 Γ β3 Γ cos ΞΈ β1 = 3 cos ΞΈ cos ΞΈ = (β1)/3 ΞΈ = cosβ1 ((βπ)/π) Therefore, the angle between π β and π β is cos-1((β1)/3)