If $a$ $a$ and $b$ $b$ are unit vectors and $θ$ $theta$ is the angle between them, express $| a - b |$ $abs(a-b)$ in terms of $θ$ $theta$ ?

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Answer 1 · 2017-08-24T09:48:05

$∣ ∣ ∣ \to a - \to b ∣ ∣ ∣ = 2 sin (\frac{θ}{2}) .$ $|veca-vecb|=2sin(theta/2).$

Suppose that, $\to a, and \to b$ $veca, and vecb$ are such unit vectors, that,

$hat{((veca, vecb))}=theta, theta in [0,pi].$

$because veca, &, vecb" are unit vectors, ":. |veca|=|vecb|=1....(0).$

We know, $(1): veca*vecb=|veca|*|vecb|*costheta,$ and,

$(2): |vecx|^2=vecx*vecx.$

$:., |veca-vecb|^2=(veca-vecb)*(veca-vecb)............[because, (2)],$

$=veca*(veca-vecb)-vecb*(veca-vecb),$

$=veca*veca-veca*vecb-vecb*veca+vecb*vecb,$

$=|veca|^2-2veca*vecb+|vecb|^2...[because, veca*vecb=vecb*veca],$

$=1-2|veca|*|vecb|*costheta+1......[because, (0), and, (1)],$

$=2-2*1*1costheta...........[because, (0)],$

$=2(1-costheta)=2(2sin^2(theta/2)).$

$rArr |veca-vecb|=2sin(theta/2).$

Enjoy Maths.!

If aaa and bbb are unit vectors and thetaθtheta is the angle between them, express abs(a-b)|a−b|abs(a-b) in terms of thetaθtheta ?