Question

Let $\overrightarrow{r}=2\hat{i}+2\hat{j}+5\hat{k}$ and A, B be the points $(1,2,5)$ and $(-1,-2,-3)$ respectively. If $\overrightarrow{BA}\times \overrightarrow{r}=4\hat{i}+6\hat{j}+2\lambda \hat{k}$ then $-\lambda =$

• A. $0.02$
• B. $-2$
• C. $2$
• D. $20$

Answer 1

Answer: (C)

$2$

We have,

$\overrightarrow{r}=2\hat{i}+2\hat{j}+5\hat{k}$

$A(1,2,5)$ and $B(-1,-2,-3)$

Now,

$\begin{array}{c}\overrightarrow{BA}=\left(1-\left(-1\right)\right)\hat{i}+\left(2-\left(-2\right)\right)\hat{j}+\left(5-\left(-3\right)\right)\hat{k}=2\hat{i}+4\hat{j}+8\hat{k}.\\ \overrightarrow{BA}\times \overrightarrow{r}=\hat{i}\left(20-16\right)-\hat{j}\left(10-16\right)+\hat{k}\left(4-8\right)=4\hat{i}+6\hat{j}-4\widehat{k.}....\left(1\right)\end{array}$

Also its given that $\overrightarrow{BA}\times \overrightarrow{r}=4\hat{i}+6\hat{j}+2\lambda \hat{k}....\left(2\right)$

By comparing equations $\left(1\right)$ and $\left(2\right)$

we get, $\lambda =-2$

Therefore $-\lambda =2$