Question

Let $\overline{p}=4\hat{\imath}-\hat{\jmath}+\hat{k}$, $\overline{q}=11\hat{\imath}-\hat{\jmath}+\hat{k}$ and $\overline{r}$ be a vector such that $(\overline{p}+\overline{q})\times\overline{r}=\overline{r}\times(-2\overline{p}+3\overline{q})$. If $(2\overline{p}+3\overline{q})\cdot\overline{r}=1670$, then $|\overline{r}|^{2}$ is equal to:

• A. 1600
• B. 1618
• C. 1627
• D. 1609

Answer 1

Answer: (B)

1618

Answer 2

The first condition $(\overline{p}+\overline{q})\times\overline{r}=\overline{r}\times(-2\overline{p}+3\overline{q})$ simplifies to $(-\overline{p}+4\overline{q})\times\overline{r}=\overline{0}$, implying $\overline{r}$ is parallel to $-\overline{p}+4\overline{q}$. Thus, $\overline{r}=k(-\overline{p}+4\overline{q})$. Calculating $-\overline{p}+4\overline{q} = 40\hat{\imath}-3\hat{\jmath}+3\hat{k}$. The second condition $(2\overline{p}+3\overline{q})\cdot\overline{r}=1670$ is used to find $k$. Calculating $2\overline{p}+3\overline{q} = 41\hat{\imath}-5\hat{\jmath}+5\hat{k}$. Substituting $\overline{r}=k(40\hat{\imath}-3\hat{\jmath}+3\hat{k})$ into the dot product gives $k(41\hat{\imath}-5\hat{\jmath}+5\hat{k})\cdot(40\hat{\imath}-3\hat{\jmath}+3\hat{k}) = 1670$, which simplifies to $k(1640+15+15)=1670$, so $1670k=1670$, leading to $k=1$. Therefore, $\overline{r}=40\hat{\imath}-3\hat{\jmath}+3\hat{k}$. Finally, $|\overline{r}|^2 = 40^2+(-3)^2+3^2 = 1600+9+9 = 1618$.