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

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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