Question

Find the vector $\overrightarrow{r}$ which satisfying the following conditions:
i) $\overrightarrow{r}$ is perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}$, where $\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k},\overrightarrow{b}=18\hat{i}-22\hat{j}-5\hat{k}$
ii) $\overrightarrow{r}$ makes an acute angle with $\hat{i}+\hat{j}+\hat{k}$
iii) $|\overrightarrow{r}|=14$

• A. $-4\hat{i}-6\hat{j}+12\hat{k}$
• B. $4\hat{i}+6\hat{j}-12\hat{k}$
• C. $-6\hat{i}-4\hat{j}+12\hat{k}$
• D. $-6\hat{i}-4\hat{j}-12\hat{k}$

Answer 1

The correct option is A $-4\hat{i}-6\hat{j}+12\hat{k}$

$\overline{r}=\lambda (\overline{a}\times \overline{b})$
$\therefore \overline{r}=\lambda (-66\overline{k}+15\overline{j}-36\overline{k}-10\overline{i}+36\overline{j}+44\overline{i})$
$=\lambda (34\overline{i}+51\overline{j}-102\overline{k})$
$=\alpha (2\overline{i}+3\overline{j}-6\overline{k})$
Now, acute angle with $\overline{i}+\overline{j}+\overline{k}$, so their dot product must be positive.
Thus,$-\alpha >0,\Rightarrow \alpha <0.$
Now the modulus is $14$, so $|\alpha |\left(\sqrt{4+9+36}\right)=14$
$\Rightarrow \alpha =-2$

Hence, option $A$.