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Question

If a unit vector ¯a makes angles π3 with ^i , π4 with ^j and an acute angle θ with ^k, find θ and hence, the component of a.

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Solution

Let us take a unit vector
a=x^i+y^j+z^k
So magnitude of a=|a|=1
Angle of a with ^i=π3
a^i=|a|icosπ3
(x^i+y^j+z^k).^i=1×1×12
(x^i+y^j+z^k).(1^i+0^j+0^k)=12
(x×1)+(y×0)+(z×0)=12
x+0+0=12
x=12
Angle of awith ^j=π4
a^j=|a|jcosπ4
(x^i+y^j+z^k).^j=1×1×12
(x^i+y^j+z^k).(0^i+1^j+0^k)=12
(x×0)+(y×1)+(z×0)=12
0+y+0=12
y=12
Angle of a with ^k=θ
a.^k=a^k×cosθ
(x^i+y^j+z^k).(0^i+0^j+1^k)=1×1×cosθ
(x×0)+(y×0)+(z×1)=cosθ
0+0+z=cosθ
z=cosθ
Now, magnitude of a=x2+y2+z2
1= (12)2+(12)2+cos2θ (As |a=1|)
1=14+12+cos2θ
1=34+cos2θ
(34+cos2θ)2=12
34+cos2θ=1
cos2θ=14
cosθ=±14
cosθ=±12
Since θ is given an acute angle
So, θ<90°
θ is in first quadrant
& cosθ is positive in first quadrant
So, cosθ=+12
θ=60°=π3
Also, z=cosθ=cos60°=12
Hence, x=12,y=12 & z=12
The required vector a is 12^i+12^j+12^k
So, components of a are 12,12 &12

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