Question : The angle between the z-axis and the vector i+j+√2k is
a) 30°
b) 45°
c) 60°
d) 90°
Doubt by Avani
Solution :
We know, angle made by a vector with x-axis, y-axis and z-axis are α, β and γ which are given by
cosα = Ax/A
cosβ = Ay/A
cosγ = Az/A
i+j+√2k
Here
Ax=1
Ay=1
Az=√2
A = √[(1)²+(1)²+(√2)²]
A = √[1+1+2]
A = √[4]
A = 2
cosγ = Az/A
cosγ = √2/2
cosγ = 1/√2
cosγ = cos45°
γ = 45°
Hence, b) 45° would be the correct option.
Alternate Method
Let
A = i+j+√2k
B = 0i+0j+k (Vector Along the z-axis)
A.B = (i+j+√2k).(0i+0j+k)
A.B = (1)(0)+(1)(0)+(√2)(1)
A.B = √2
A = √[(1)²+(1)²+(√2)²]
A = √[1+1+2]
A = √[4]
A = 2
B = 1
We know,
cosθ = A.B/AB
cosθ = √2 / [(2)(1)]
cosθ = √2/2
cosθ = 1/√2
cosθ = cos45°
Hence, b) 45° would be the correct option.
Similar Question :
If P=4i-2j+6k and Q=i-2j-3k, then the angle which P+Q makes with x-axis is
a) cos-1[3/√50]
b) cos-1[4/√50]
c) cos-1[5/√50]
d) cos-1[12/√50]
Ans : c) cos-1[5/√50]