Vector Algebra
Finding a vector satisfying a cross product condition; maximum of |c|²
nta_pyq_2025_apr
Grade 12
Question:
Let $\overrightarrow{a}=2\hat{i}-\hat{j}+3\hat{k}$, $\overrightarrow{b}=3\hat{i}-5\hat{j}+\hat{k}$ and $\overrightarrow{c}$ be a vector such that $\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{c}\times\overrightarrow{b}$ and $(\overrightarrow{a}+\overrightarrow{c})\cdot(\overrightarrow{b}+\overrightarrow{c})=168$. Then the maximum value of $|\overrightarrow{c}|^2$ is:
Step-by-Step Solution
Key Concept: $\vec{a}\times\vec{c}=\vec{c}\times\vec{b}$ implies $(\vec{a}+\vec{b})\times\vec{c}=\vec{0}$, so $\vec{c}=\lambda(\vec{a}+\vec{b})$; substitute into the dot product condition to get a quadratic in $\lambda$, then maximise $|\vec{c}|^2=77\lambda^2$.
$(\vec{a}+\vec{b})\times\vec{c}=\vec{0} \Rightarrow \vec{c}=\lambda(\vec{a}+\vec{b})=\lambda(5\hat{i}-6\hat{j}+4\hat{k})$.
$|\vec{c}|^2=77\lambda^2$.
$(\vec{a}+\vec{c})\cdot(\vec{b}+\vec{c})=\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}+\vec{c}\cdot\vec{b}+|\vec{c}|^2=168$.
$\vec{a}\cdot\vec{b}=6+5+3=14$. $\vec{c}\cdot(\vec{a}+\vec{b})=\lambda|\vec{a}+\vec{b}|^2=77\lambda$. So $\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}=77\lambda$.
$14+77\lambda+77\lambda^2=168 \Rightarrow \lambda^2+\lambda-2=0 \Rightarrow \lambda=1$ or $\lambda=-2$.
Maximum $|\vec{c}|^2=77\times4=308$ (at $\lambda=-2$).
Correct Answer: 4