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)
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
Transcript
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