Question

$\overline{\alpha },\overline{\beta }$ are two unit vectors $\overline{r}$ is a vector such that $\overline{r}.\overline{\alpha }=0$ and $\sqrt{2}(\overline{r}\times \overline{\beta })=3(\overline{r}\times \overline{\alpha })\overline{-\beta }$, then $\frac{1}{{\overline{\left|r\right|}}^2}$ equals to ?

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is D $7$

Given $\overline{\alpha },\overline{\beta }$ are two unit vectors and $\overline{r}\cdot \overline{\alpha }=0$

Given, $\sqrt{2}\left(\overline{r}\times \overline{\beta }\right)=3\left(\overline{r}\times \overline{\beta }\right)-\overline{\beta }$

Applying dot product w.r.t $\overline{r}$ on both side

$\sqrt{2}\overline{r}\cdot \left(\overline{r}\times \overline{\beta }\right)=3\left(\overline{r}\times \overline{\alpha }\right)-\overline{r}\cdot \overline{\beta }$

$\overline{r}\overline{\cdot \beta }=0$

$0=0-\overline{r}\overline{\cdot \beta }$

$\left|\sqrt{2}\left(\overline{r}\times \overline{\beta }\right)+\overline{\beta }\right|=\left|3\left(\overline{r}\times \overline{\alpha }\right)\right|$

$\sqrt{{\left(\sqrt{2}\left|r\right|\left|1\right|\left|1\right|\right)}^2+1^2+2\cdot \sqrt{2}\cdot \cos \left(\overline{\beta },\left(\overline{r}\times \overline{\beta }\right)\right)}=3\left|r\right|$

since $\left(\overline{r}\times \overline{\beta }\right)$ is perpendicular to $\overline{\beta },\cos \left(\overline{\beta },\left(\overline{r}\times \overline{\beta }\right)is0\right)$

$\sqrt{2{\left|r\right|}^2+1+0}=3\left|r\right|$

squaring on both side

$2{\left|r\right|}^2+1=9{\left|r\right|}^2$

$1=7{\left|r\right|}^2$

${\left|r\right|}^2=\frac{1}{7}$

$so,\frac{1}{{\left|r\right|}^2}=\frac{\frac{1}{1}}{7}=7$

so answer is 7