Vector Algebra
Finding a vector from cross and dot conditions
nta_pyq_2023_jan
Grade 12
Question:
Let $\vec{a} = 2\hat{i}-7\hat{j}+5\hat{k}$, $\vec{b} = \hat{i}+\hat{k}$ and $\vec{c} = \hat{i}+2\hat{j}-3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}\times\vec{a} = \vec{c}\times\vec{a}$ and $\vec{r}\cdot\vec{b}=0$, then $|\vec{r}|$ is equal to:
\frac{11}{7}\sqrt{2}
\frac{11}{7}
\frac{11}{5}\sqrt{2}
\frac{\sqrt{914}}{7}
Step-by-Step Solution
Key Concept: From $\vec{r}\times\vec{a} = \vec{c}\times\vec{a}$: $\vec{r} = \vec{c}+\lambda\vec{a}$. Use $\vec{r}\cdot\vec{b}=0$ to find $\lambda$.
$\lambda=2/7$. $\vec{r} = (1+4/7)\hat{i}+(-2-2)\hat{j}+(-3+10/7)\hat{k} = (11/7)\hat{i}-4\hat{j}+(-11/7)\hat{k}$. Wait: $\vec{r} = \hat{i}+2\hat{j}-3\hat{k}+(2/7)(2\hat{i}-7\hat{j}+5\hat{k}) = (11/7)\hat{i}+0\hat{j}+(-11/7)\hat{k}$. $|\vec{r}| = \frac{11}{7}\sqrt{2}$. Answer: (1)
Correct Answer: $\frac{11}{7}\sqrt{2}$