If a Μ and b Μ are unit vectors, then prove that
|a Μ+b Μ |= 2πππ θ/2, where π is the angle between them.
This question is similar to MIsc 17 (MCQ) - Chapter 10 Class 12 - Vector Algebra
CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
Question 1 (Choice 2)
Question 2 Important
Question 3 You are here
Question 4 Important
Question 5
Question 6 Important
Question 7 Important
Question 8 (Choice 1)
Question 8 (Choice 2)
Question 9 Important
Question 10 (Choice 1)
Question 10 (Choice 2)
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Question 12 (Choice 1)
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CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
Last updated at Dec. 14, 2024 by Teachoo
This question is similar to MIsc 17 (MCQ) - Chapter 10 Class 12 - Vector Algebra
Question 3 If π Μ and π Μ are unit vectors, then prove that |π Μ+π Μ |= 2πππ π/2, where π is the angle between them. Given π Μ & π Μ are unit vectors, So, |π Μ | = 1 & |π Μ | = 1 We need to find |π Μ+π Μ | Letβs find |π Μ+π Μ |^π instead Now, |π Μ+π Μ |^π=(π Μ+π Μ ).(π Μ+π Μ ) |π Μ+π Μ |^2=π Μ.(π Μ+π Μ )+π Μ.(π Μ+π Μ ) (As |π β |^2 = π β.π β) |π Μ+π Μ |^2=π Μ.π Μ+π Μ.π Μ+π Μ.π Μ+π Μ.π Μ |π Μ+π Μ |^2=|π Μ |^π+π Μ.π Μ+π Μ.π Μ+|π Μ |^π |π Μ+π Μ |^2=π^π+π Μ.π Μ+π Μ.π Μ+π^π |π Μ+π Μ |^2=π+π Μ.π Μ+π Μ.π Μ |π Μ+π Μ |^2=2+π Μ.π Μ+π Μ.π Μ |π Μ+π Μ |^2=2+2π Μ.π Μ |π Μ+π Μ |^2=2+2|π Μ|.|π Μ | ππ¨π¬β‘π½ |π Μ+π Μ |^2=2+2 Γ 1 Γ 1 πππ β‘π |π Μ+π Μ |^2=2+2 πππ β‘π |π Μ+π Μ |^2=2(π+πππβ‘π½) Using cos 2ΞΈ = 2cos2 ΞΈ β (As |π β |^2 = π β.π β) (As |π Μ |=1 & |π Μ |=1) (As |π Μ |=1 & |π Μ |=1) |π Μ+π Μ |^2=2 Γ ππππ^π π½/π |π Μ+π Μ |^2=4πππ ^2 π/2 |π Μ+π Μ |^2=(2πππ π/2)^2 Taking square root both sides |π Μ+π Μ |=ππππ π½/π Hence proved