Vectors & 3D Geometry
Vector equation; magnitude of sum
MMTS_Full_Test_06
Grade 12
Question:
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three non-zero vectors satisfying $\vec{a}=\vec{b}\times\vec{c}+2\vec{b}$, where $|\vec{b}|=|\vec{c}|=2$ and $|\vec{a}|\leq4$. The sum of possible values of $|2\vec{a}+\vec{b}+\vec{c}|$ is
(A) 8
(B) 12
(C) 20
(D) 32
Step-by-Step Solution
Key Concept: Take dot product with $\vec{b}$: $\vec{a}\cdot\vec{b}=2|\vec{b}|^2=8$. Use $|\vec{a}|\leq4$ to find $|\vec{a}|=4$ and $\vec{a}\parallel\vec{b}$ ($\theta=0$). Then $\vec{b}\times\vec{c}=0$, so $\vec{b}=\vec{c}$ or $\vec{b}=-\vec{c}$.
$\vec{a}=2\vec{b}$. $|2\vec{a}+\vec{b}+\vec{c}|=|5\vec{b}+\vec{c}|$: if $\vec{b}=\vec{c}$: $|6\vec{b}|=12$; if $\vec{b}=-\vec{c}$: $|4\vec{b}|=8$. Sum$=20$.
Correct Answer: (C) 20