Vector Algebra
Vector $\vec{r}$ from Cross & Dot Conditions — Finding $|\vec{b}+\vec{c}|^2$
nta_pyq_2024_apr
Grade 12
Question:
Let $\vec{a}=9\hat{i}-13\hat{j}+25\hat{k}$, $\vec{b}=3\hat{i}+7\hat{j}-13\hat{k}$ and $\vec{c}=17\hat{i}-2\hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}\times\vec{a}=(\vec{b}+\vec{c})\times\vec{a}$ and $\vec{r}\cdot(\vec{b}-\vec{c})=0$, then $\dfrac{|593\vec{r}+67\vec{a}|^2}{(593)^2}$ is equal to _____
Step-by-Step Solution
Key Concept: $\vec{r}\times\vec{a}=(\vec{b}+\vec{c})\times\vec{a}\Rightarrow(\vec{r}-(\vec{b}+\vec{c}))\times\vec{a}=0\Rightarrow\vec{r}=\vec{b}+\vec{c}+\lambda\vec{a}$. $\vec{r}\cdot(\vec{b}-\vec{c})=0\Rightarrow(\vec{b}+\vec{c}+\lambda\vec{a})\cdot(\vec{b}-\vec{c})=0$.
$\lambda=-67/593$. $|\vec{b}+\vec{c}|^2=569$.
Correct Answer: 569