NEET Practice Questions (MCQs) with Answers & Solutions

$$\begin{gathered} The unit vector=\frac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|} \\ =\frac{5\hat{i}-12\hat{j}}{\sqrt{{\left(5\right)}^2+{\left(12\right)}^2}} \\ =\frac{1}{13}(5\hat{i}-12\hat{j}) \end{gathered}$$

$$\begin{gathered} Initial position (x_1, y_1) = (1,-2) \\ Final position (x_2, y_2) = (4,2) \\ Displacement = Final position -Initial position \\ =( x_2 -x_1, y_2 -y_1) \\ =(3, 4) \\ = 3 \hat{i} + 4 \hat{j} \\ Direction of velocity is same as diplacement \\ So unit vector along displacement = \frac{Vector}{Magnitude} \\ =\frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^2+4^2}}=\frac{3 \hat{i} + 4 \hat{j}}{5} \\ Velocity = Magnitude of velocity \times Unit Vector along velocity \\ =10 \times \left(\frac{3 \hat{i} + 4 \hat{j}}{5}\right) \\ =6 \hat{i} + 8 \hat{j} \end{gathered}$$

$$\begin{gathered} If vector \overrightarrow{A } is \perp to \overrightarrow{B, } \\ \overrightarrow{B }.\overrightarrow{A }=0 \\ Checking all options \\ a) (\hat{i} - \hat{j} -\hat{k}). (\hat{i} + \hat{j} +\hat{k}) = 1-1-1 =-1 (Not perpendicular\left) \\ b\right) (\hat{i} - \hat{j} -\hat{k}). (\widehat{-i} + \hat{j} +\hat{k}) = -1-1-1 =-3 (Not perpendicular) \\ c) (\hat{i} - \hat{j} -\hat{k}). (\hat{i} + \hat{j} -\hat{k}) = -1-1+1 =1 (Not perpendicular) \\ \mathit{S}\mathit{o}\mathbf{ }\mathbf{,}\mathbf{ }\mathit{n}\mathit{o}\mathbf{ }\mathit{o}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{m}\mathit{a}\mathit{t}\mathit{c}\mathit{h}\mathit{e}\mathit{s}\mathbf{.}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\mathbf{ }\mathit{i}\mathit{s}\mathbf{ }\mathit{o}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{D} \end{gathered}$$

(A)

$$\begin{gathered} \mathrm{Net} \mathrm{force}= {\overrightarrow{\mathrm{F}}}_1+{\overrightarrow{\mathrm{F}}}_2 \\ \left({\overrightarrow{\mathrm{F}}}_{\mathrm{net}}\right)= \left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)+\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}\right) \\ = 3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \mathrm{Displacement} ∆\overrightarrow{\mathrm{r}}= {\overrightarrow{\mathrm{r}}}_2-{\overrightarrow{\mathrm{r}}}_1 \\ = 6\hat{\mathrm{i}}+8\hat{\mathrm{j}}+7\hat{\mathrm{k}} \\ \mathrm{Work} \mathrm{done}= \mathrm{Force}. \mathrm{Displacement} \\ = {\overrightarrow{\mathrm{F}}}_{\mathrm{net}}.∆\overrightarrow{\mathrm{r}} \\ = \left(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}\right).\left(6\hat{\mathrm{i}}+8\hat{\mathrm{j}}+7\hat{\mathrm{k}}\right) \\ = 18-16+7 \\ = 9 \mathrm{J} \end{gathered}$$

(B)

$$\begin{gathered} \overrightarrow{A}.\overrightarrow{B}=2\times 7+1\times (-5)+3\times (-3) \\ =14-5-9=0 \\ \Rightarrow perpendicular \end{gathered}$$

$$\begin{gathered} \left(D\right) \\ For \overrightarrow{P} \perp \stackrel{ }{ \overrightarrow{Q}} , \\ Dot product \overrightarrow{P}.\overrightarrow{Q} =0 \\ (2\hat{i} +bj +2\hat{k}). (\hat{i} +j +\hat{k}) = 0 \\ 2 +b +2 = 0 \\ b =-4 \end{gathered}$$

(B)$$\begin{gathered} (4\hat{i} - 3\hat{j}).7\widehat{k } = 0 \\ So \widehat{k } is the required vector \end{gathered}$$

The angle that a vector makes with the y-axis is given by tan^-1(y/x). Here, the vector A = 2i + 3j has x = 2 and y = 3. Therefore, the angle it makes with the y-axis is tan^-1(3/2).