NIMCET Previous Year Questions (PYQs) – Vector – Page 1 of 7

Given: $\alpha (2\overrightarrow{a}-\overrightarrow{b})+\beta (\overrightarrow{a}+2\overrightarrow{b})=8\overrightarrow{b}-\overrightarrow{a}$ for all vectors $\overrightarrow{a},\overrightarrow{b}$.

Combine like terms: $(2\alpha +\beta )\overrightarrow{a}+(-\alpha +2\beta )\overrightarrow{b}=-\overrightarrow{a}+8\overrightarrow{b}$.

Equate coefficients:
$2\alpha +\beta =-1$,
$-\alpha +2\beta =8$.

Solve:
From $2\alpha +\beta =-1\Rightarrow \beta =-1-2\alpha$.
Substitute in $-\alpha +2\beta =8$:
$-\alpha +2(-1-2\alpha )=8\Rightarrow -\alpha -2-4\alpha =8\Rightarrow -5\alpha =10\Rightarrow \alpha =-2$.
Then $\beta =-1-2(-2)=3$.

Answer: $\alpha =-2,\beta =3$.

Given: Force magnitude $=78$ (grams), point of application $\mathbf{r}=(2,3,5)$, force direction ratios $(2,2,1)$. Line (about which moment is required) is along $(12,3,4)$ through the origin.

Force vector: Unit along $(2,2,1)$ is ${\displaystyle \frac{1}{\sqrt{2^2+2^2+1^2}}}(2,2,1)={\displaystyle \frac{1}{3}}(2,2,1)$. Hence $$\mathbf{F}=78\cdot \frac{1}{3}(2,2,1)=(52,52,26).$$

Torque about origin: $$\mathit{\tau }=\mathbf{r}\times \mathbf{F}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 3& 5\\ 52& 52& 26\end{array}\right|=(-182,208,-52).$$

Moment about the line with direction $\mathbf{L}=(12,3,4)$: Let $\hat{\ell }={\displaystyle \frac{\mathbf{L}}{\Vert \mathbf{L}\Vert }}$. The required magnitude is the component of $\mathit{\tau }$ along $\hat{\ell }$: $$M=|\mathit{\tau }\cdot \hat{\ell }|=\frac{|\mathit{\tau }\cdot \mathbf{L}|}{\Vert \mathbf{L}\Vert }=\frac{|(-182,208,-52)\cdot (12,3,4)|}{\sqrt{{12}^2+3^2+4^2}}=\frac{|-1768|}{13}=136.$$

Answer: $\boxed{{\displaystyle 136}}$ (in gram–unit of length).