Vector Algebra Part 3

Concept:Use dot product expansions to simplify the sum of squared magnitudes and deduce that the vectors sum to zero, then compute individual dot products.Explanation:Given $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$.Expand $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$:$(|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b})+(|\vec{b}|^2+|\vec{c}|^2-2\vec{b}\cdot\vec{c})+(|\vec{c}|^2+|\vec{a}|^2-2\vec{c}\cdot\vec{a})=9$.Since $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$, we get $6-2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=9$.Thus $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}=-\frac{3}{2}$.Now $|\vec{a}+\vec{b}+\vec{c}|^2=3+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=3+2(-\frac{3}{2})=0$, so $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.From this, $\vec{a}+\vec{b}=-\vec{c}$, so $|\vec{a}+\vec{b}|^2=|\vec{c}|^2=1$.Expanding: $|\vec{a}|^2+|\vec{b}|^2+2\vec{a}\cdot\vec{b}=1+1+2\vec{a}\cdot\vec{b}=1$, giving $\vec{a}\cdot\vec{b}=-\frac{1}{2}$.Similarly, $\vec{b}\cdot\vec{c}=-\frac{1}{2}$ and $\vec{c}\cdot\vec{a}=-\frac{1}{2}$.Now use $|2\vec{a}+k\vec{b}+k\vec{c}|=3$. Square both sides:$|2\vec{a}+k\vec{b}+k\vec{c}|^2=9$.Expand: $4|\vec{a}|^2+k^2|\vec{b}|^2+k^2|\vec{c}|^2+4k\vec{a}\cdot\vec{b}+2k^2\vec{b}\cdot\vec{c}+4k\vec{a}\cdot\vec{c}=9$.Substitute norms and dot products: $4(1)+k^2(1)+k^2(1)+4k(-\frac{1}{2})+2k^2(-\frac{1}{2})+4k(-\frac{1}{2})=9$.Simplify: $4+2k^2-2k-k^2-2k=9$, i.e. $k^2-4k+4=9$? Wait check: $4+2k^2 -2k -k^2 -2k = 4+(2k^2-k^2)+(-2k-2k)=4+k^2-4k$.So $4+k^2-4k=9$, thus $k^2-4k-5=0$.Factor: $(k-5)(k+1)=0$, so $k=5$ or $k=-1$.Positive value is $k=5$.Answer:Option B. 5