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Ex 10.4, 3 - If a unit vector a makes angles pi/3 with i, pi/4

Ex 10.4, 3 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.4, 3 - Chapter 10 Class 12 Vector Algebra - Part 3 Ex 10.4, 3 - Chapter 10 Class 12 Vector Algebra - Part 4 Ex 10.4, 3 - Chapter 10 Class 12 Vector Algebra - Part 5


Transcript

Ex 10.4, 3 If a unit vector 𝑎 ⃗ makes angles 𝜋/3 with 𝑖 ̂, 𝜋/4 , with 𝑗 ̂ & an acute angle θ with 𝑘 ̂ , then find θ and hence, the components of 𝑎 ⃗ . Let us take a unit vector 𝑎 ⃗ = 𝑥𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂ So, magnitude of 𝑎 ⃗ = |𝑎 ⃗ | = 1 Angle of 𝒂 ⃗ with 𝒊 ̂ = 𝝅/𝟑 𝑎 ⃗ . 𝑖 ̂ = |𝑎 ⃗ ||𝑖 ̂ | cos 𝜋/3 (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂). 𝑖 ̂ = 1 × 1 × 1/2 (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂). (1𝑖 ̂ + 0𝑗 ̂ + 0𝑘 ̂) = 1/2 (x × 1) + (y × 0) + (z × 0) = 1/2 x + 0 + 0 = 1/2 x = 𝟏/𝟐 Angle of 𝒂 ⃗ with 𝒋 ̂ = 𝝅/𝟒 𝑎 ⃗ . 𝑗 ̂ = |𝑎 ⃗ ||𝑗 ̂ | cos 𝜋/4 (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂). 𝑗 ̂ = 1 × 1 × 1/√2 (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂). (0𝑖 ̂ + 1𝑗 ̂ + 0𝑘 ̂) = 1/√2 (x × 0) + (y × 1) + (z × 0) = 1/√2 0 + y + 0 = 1/√2 y = 𝟏/√𝟐 Also, Angle of 𝑎 ⃗ with 𝑘 ̂ = θ 𝑎 ⃗. 𝑘 ̂ = |𝑎 ⃗ ||𝑘 ̂ |×cos⁡"θ" (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂). (0𝑖 ̂ + .0𝑗 ̂ + 1𝑘 ̂) = 1 × 1 × cos θ (x × 0) + (y × 0) + (z × 1) = cosθ 0 + 0 + z = cos θ z = cos θ Now, Magnitude of 𝑎 ⃗ = √(𝑥^2+𝑦2+𝑧2) 1 = √((1/2)^2+(1/√2)^2+𝑐𝑜𝑠2"θ" ) 1 = √(1/4+1/2+𝑐𝑜𝑠2"θ" ) 1 = √(3/4+𝑐𝑜𝑠2"θ" ) √(3/4+𝑐𝑜𝑠2"θ" ) = 1 (√(3/4+𝑐𝑜𝑠2"θ" ))^2 = 12 3/4 + 𝑐𝑜𝑠2" θ" = 1 𝑐𝑜𝑠2 "θ" = 1 − 3/4 𝑐𝑜𝑠2" θ" = 1/4 cos⁡"θ" = ± √(1/4) cos⁡"θ" = ± 1/2 Since θ is given an acute angle So, θ < 90° ∴ θ is in 1st quadrant And, cos θ is positive in 1st quadrant= So, cos θ = 1/2 ∴ θ = 60° = 𝝅/𝟑 Also, z = cos θ = cos 60° = 𝟏/𝟐 Hence x = 1/2 , y = 1/√2 & z = 1/2 The required vector 𝑎 ⃗ is 1/2 𝑖 ̂ + 1/√2 𝑗 ̂ + 1/2 𝑘 ̂ So, components of 𝑎 ⃗ are 𝟏/𝟐 , 𝟏/√𝟐 & 𝟏/𝟐

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.