Vector Algebra

$ \begin{aligned} &\text { According to question, }\\ &\begin{aligned} A \cos \theta & =2 B \cos \theta \\ A & =2 B \\ \Rightarrow \quad \frac{A}{B} & =\frac{2}{1} \Rightarrow A: B=2: 1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text {Given, }\\ &\begin{aligned} & \text { Force }=\mathbf{F}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\ & \text { Displacement }=\mathbf{d}=5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \qquad \mathbf{F} \cdot \mathbf{d}=F d \cos \theta \quad \text { [Dot product] } \\ & \cos \theta=\frac{\mathbf{F} \cdot \mathbf{d}}{F \times d}=\frac{15+16-15}{50}=0.32 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \cos \theta & =0.32 \\ \theta & =\cos ^{-1}(0.32) \end{aligned}\\ &\text { Hence, correct option is (b). } \end{aligned} $

Displacement, $\mathrm{s}=(10 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) \mathrm{cm}$

$ \begin{aligned} &\begin{aligned} & r_x=\sqrt{2} \cos 45^{\circ}=1 \mathrm{~cm} \\ & r_y=\sqrt{2} \sin 45^{\circ}=1 \mathrm{~cm} \end{aligned}\\ &\text { Now, dot product of } \mathbf{s} \text { and } \mathbf{r}\\ &\begin{aligned} \mathbf{s} \cdot \mathbf{r} & =(10 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \\ & =(10+20)(\because \hat{\mathbf{i}} \cdot \hat{\mathbf{i}}=1, \hat{\mathbf{j}} \cdot \hat{\mathbf{j}}=1, \hat{\mathbf{i}} \cdot \hat{\mathbf{j}}=0) \\ & =30 \mathrm{~cm} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } & \text { Given, } \mathbf{r}_1 & =2 \hat{\mathbf{x}} \Rightarrow \mathbf{r}_2=2 \hat{\mathbf{y}} \\ & \therefore & \mathbf{r}_1+\mathbf{r}_2 & =2 \hat{\mathbf{x}}+2 \hat{\mathbf{y}} \\ & \therefore & \left|\mathbf{r}_1+\mathbf{r}_2\right| & =\sqrt{2^2+2^2}=\sqrt{8}=2 \sqrt{2} \end{aligned} $

Given that,

$ \begin{align*} & \mathbf{A}_1+\mathbf{A}_2=5 \mathbf{A}_3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \mathbf{A}_1-\mathbf{A}_2=3 \mathbf{A}_3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

On adding Eqs. (i) and (ii), we get

$ \quad \begin{aligned} 2 \mathbf{A}_1 & =8 \mathbf{A}_3 \\\Rightarrow \mathbf{A}_1 & =4 \mathbf{A}_3 \\ \mathbf{A}_1 & =4(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \\ & {\left[\because \mathbf{A}_3=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} \text { (given) }\right] } \\ \mathbf{A}_1= & 8 \hat{\mathbf{i}}+16 \hat{\mathbf{j}} \end{aligned} $

On subtracting Eqs. (ii) from (i), we get

$ \begin{aligned} 2 \mathbf{A}_2 & =2 \mathbf{A}_3 \\\Rightarrow \mathbf{A}_2 & =\mathbf{A}_3=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} \\ \frac{\left|\mathbf{A}_1\right|}{\left|\mathbf{A}_2\right|} & =\frac{|8 \hat{\mathbf{i}}+16 \hat{\mathbf{j}}|}{|2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}|} \\ & =\frac{\sqrt{(8)^2+(16)^2}}{(2)^2+(4)^2} \\ & =\frac{\sqrt{64+256}}{\sqrt{4+16}} \\ & =\frac{\sqrt{320}}{\sqrt{20}}=\sqrt{16}=4 \end{aligned} $

$ \begin{aligned} &\text { Given, unit vector is } 0.5 \hat{\mathbf{i}}+0.8 \hat{\mathbf{j}}+c \hat{\mathbf{k}}\\ &\begin{gathered} =\frac{1}{2} \hat{\mathbf{i}}+\frac{4}{5} \hat{\mathbf{j}}+c \hat{\mathbf{k}} \\ \Rightarrow \sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{4}{5}\right)^2+c^2}=1 \\ \Rightarrow \quad \frac{1}{4}+\frac{16}{25}+c^2=1 \Rightarrow c^2=\frac{11}{100} \Rightarrow c=\sqrt{0.11} \end{gathered} \end{aligned} $