If a, b, c are unit vectors satisfying $|a-b|^2+|b-c|^2+|c-a|^2=9$ then find the value of $|2a+5b+5c|$
Options are:
A: $1$,
B: $2$,
C: $3$,
D: $4$
considering $|2a+5b+5c|^2$ as $k$ then
$54+20(a*b)+50(b*c)+20(c*a)=k$ and we know the value of $(a*b+b*c+c*a)$
where * is dot product . How to eliminate extra $(b*c)$
Let $A$ denote the angle between vectors $a$ and $b$, $B$ the angle between $b$ and $c$ and $C$ the angle between $c$ and $a$. By suitably renaming the vectors if necessary, we get the relation $A+B+C = 2 \pi$.
From the given equation we get $-2(a.b+b.c+c.a) = 9$ or,
$\cos A + \cos B + \cos C = -3/2$
Now, we need to show that this is possible iff $A=B=C=2 \pi /3$
Let $A+B=2P$ and $A-B=2Q$, Then $\cos (P+Q) + \cos (P-Q) + \cos (2 \pi - 2P) = -3/2$
Using standard trigonometric identities and simplifying, we get a quadratic in $\cos P$ as follows $4 \cos ^2 P + 4 \cos Q \cos P + 1 = 0$, which has a real solution iff $\cos Q = 1$ i.e. $Q = 0$ or $2 \pi$. Its easy to show that $0$ is the only possible solution and hence $A=B$. Similarly, $B = C$. So, vectors $a, b, c$ are unit vectors with their end points forming an equilateral triangle.
Let $a = \vec{i}, b = -\frac{1}{2} \vec{i} +\frac{\sqrt{3}}{2} \vec{j}, c = -\frac{1}{2} \vec{i} -\frac{\sqrt{3}}{2} \vec{j}$
$|2a+5b+5c| = |-3 \vec{i}| = 3$