Previous Year NEET Question- Laws Of Motion L-4

Question: If $\vec{F}=2 \hat{i}+\hat{j}-\widehat{k}$ and $\vec{r}=3 \hat{i}+2 \hat{j}-2 \widehat{k}$, then the scalar and vector products of $\vec{F}$ and $\vec{r}$ have the magnitudes respectively as#

A) 10,2

B) $5, \sqrt{3}$

C) $4, \sqrt{5}$

D) $10, \sqrt{2}$

Answer: $10, \sqrt{2}$#

Solution:#

Given, $\vec{F}=2 \hat{i}+\hat{j}-\widehat{k}$ and $\vec{r}=3 \hat{i}+2 \hat{j}-2 \widehat{k}$ $$ \begin{aligned} & \vec{F} \cdot \vec{r}=2(3)+2(1)+(-2)(-1) \ & =6+2+2=10 \end{aligned} $$ $$ \vec{F} \times \vec{r}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \widehat{k} \ 2 & 1 & -1 \ 3 & 2 & -2 \end{array}\right| $$ $$ =\hat{i}(-2+2)-\hat{j}(-4+3)+\widehat{k}(4-3) $$ $$ \Rightarrow \vec{F} \times \vec{r}=\hat{j}+\widehat{k} $$ $$ |\vec{F} \times \vec{r}|=\sqrt{1^2+1^2}=\sqrt{2} $$