Question 123 of 379MediumMCQ
For three unit vectors a,b,c satisfying ∣a−b∣2+∣b−c∣2+∣c−a∣2=9 and ∣2a+kb+kc∣=3, the positive value of k is: Solutions · 1
Official solution
Expanding the first condition: 6+2(a⋅b+b⋅c+c⋅a)=9⇒a⋅b+b⋅c+c⋅a=−23.
Also a⋅b=b⋅c=c⋅a=−21 each (from ∣a+b+c∣2=3+2(−23)=0, and symmetric pairwise arguments each giving −21).
\[|2\vec a+k\vec b+k\vec c|^2 = 4+2k^2+8k\left(-\frac12\right)+2k^2\left(-\frac12\right) = 9\]
\[4+2k^2-4k-k^2 = 9 \implies k^2-4k-5=0 \implies (k-5)(k+1)=0 \implies k=5,-1\]
Positive value: k=5. Answer the question to unlock