268983 A man travels $\mathbf{1}$ mile due east, then 5 miles due south, then 2 miles due east and finally 9 miles due north. His displacement is

268958 The vector sum of two vectors of magnitudes 10 units and 15 units can never be

268956 If $\overrightarrow{A}=3\hat{i}-4\hat{j}$ and $\overrightarrow{B}=-\hat{i}-4\hat{j}$, calculate the direction of $\overrightarrow{A}-\overrightarrow{B}$.

268957 The resultant of the forces ${\overrightarrow{F}}_1=4\hat{i}-3\hat{j}$ and $\overrightarrow{F_2}=6\hat{i}+8\hat{j}$ is

268983 A man travels $\mathbf{1}$ mile due east, then 5 miles due south, then 2 miles due east and finally 9 miles due north. His displacement is

268958 The vector sum of two vectors of magnitudes 10 units and 15 units can never be

268959 A car moves $40\mathrm{m}$ due east and turns towards north and moves $30\mathrm{m}$ then turns ${45}^{\circ }$ east of north and moves $20\sqrt{2}\mathrm{m}$. The net displacement of car is ( east is taken positive $x$-axis, North as positive $\mathbf{y}$-axis)

268956 If $\overrightarrow{A}=3\hat{i}-4\hat{j}$ and $\overrightarrow{B}=-\hat{i}-4\hat{j}$, calculate the direction of $\overrightarrow{A}-\overrightarrow{B}$.

268957 The resultant of the forces ${\overrightarrow{F}}_1=4\hat{i}-3\hat{j}$ and $\overrightarrow{F_2}=6\hat{i}+8\hat{j}$ is

268983 A man travels $\mathbf{1}$ mile due east, then 5 miles due south, then 2 miles due east and finally 9 miles due north. His displacement is

268958 The vector sum of two vectors of magnitudes 10 units and 15 units can never be

268959 A car moves $40\mathrm{m}$ due east and turns towards north and moves $30\mathrm{m}$ then turns ${45}^{\circ }$ east of north and moves $20\sqrt{2}\mathrm{m}$. The net displacement of car is ( east is taken positive $x$-axis, North as positive $\mathbf{y}$-axis)

268956 If $\overrightarrow{A}=3\hat{i}-4\hat{j}$ and $\overrightarrow{B}=-\hat{i}-4\hat{j}$, calculate the direction of $\overrightarrow{A}-\overrightarrow{B}$.

268957 The resultant of the forces ${\overrightarrow{F}}_1=4\hat{i}-3\hat{j}$ and $\overrightarrow{F_2}=6\hat{i}+8\hat{j}$ is

268983 A man travels $\mathbf{1}$ mile due east, then 5 miles due south, then 2 miles due east and finally 9 miles due north. His displacement is

268958 The vector sum of two vectors of magnitudes 10 units and 15 units can never be

268959 A car moves $40\mathrm{m}$ due east and turns towards north and moves $30\mathrm{m}$ then turns ${45}^{\circ }$ east of north and moves $20\sqrt{2}\mathrm{m}$. The net displacement of car is ( east is taken positive $x$-axis, North as positive $\mathbf{y}$-axis)

268956 If $\overrightarrow{A}=3\hat{i}-4\hat{j}$ and $\overrightarrow{B}=-\hat{i}-4\hat{j}$, calculate the direction of $\overrightarrow{A}-\overrightarrow{B}$.

268957 The resultant of the forces ${\overrightarrow{F}}_1=4\hat{i}-3\hat{j}$ and $\overrightarrow{F_2}=6\hat{i}+8\hat{j}$ is

268983 A man travels $\mathbf{1}$ mile due east, then 5 miles due south, then 2 miles due east and finally 9 miles due north. His displacement is

268958 The vector sum of two vectors of magnitudes 10 units and 15 units can never be

268959 A car moves $40\mathrm{m}$ due east and turns towards north and moves $30\mathrm{m}$ then turns ${45}^{\circ }$ east of north and moves $20\sqrt{2}\mathrm{m}$. The net displacement of car is ( east is taken positive $x$-axis, North as positive $\mathbf{y}$-axis)

268956 If $\overrightarrow{A}=3\hat{i}-4\hat{j}$ and $\overrightarrow{B}=-\hat{i}-4\hat{j}$, calculate the direction of $\overrightarrow{A}-\overrightarrow{B}$.

268957 The resultant of the forces ${\overrightarrow{F}}_1=4\hat{i}-3\hat{j}$ and $\overrightarrow{F_2}=6\hat{i}+8\hat{j}$ is