BM Sharma Solutions for Chapter: Vectors, Exercise 3: DPP

A body is at rest under the action of three forces, two of which are ${\overrightarrow{F}}_1=4\hat{\mathrm{i}}$ and ${\overrightarrow{F}}_2=6\hat{\mathrm{j}}$. The third force is

The vector which must be added to the vectors $\hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and $3\hat{\mathrm{i}}+6\hat{\mathrm{j}}-7\hat{\mathrm{k}}$, so that the resultant vector is a unit vector along the $y-$axis is

If $\overrightarrow{A}=3\hat{\mathrm{i}}+4\hat{\mathrm{j}}$ and $\overrightarrow{B}=7\hat{\mathrm{i}}+24\hat{\mathrm{j}}$, then the vector having the same magnitude as $\overrightarrow{B}$ and parallel to the $\overrightarrow{A}$ is:

The unit vector parallel to the resultant of the vectors $\overrightarrow{A}=4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+6\hat{\mathrm{k}}$ and $\overrightarrow{B}=-\hat{\mathrm{i}}+3\hat{\mathrm{j}}-8\hat{\mathrm{k}}$ is

The position vector of a particle is determined by the expression $\overrightarrow{r}=3t^2\hat{\mathrm{i}}+4t^2\hat{\mathrm{j}}+7\hat{\mathrm{k}}$. The distance traversed in the first $10 \mathrm{s}$ is

A unit vector parallel to the resultant of vectors $\overrightarrow{A}=4\hat{\mathrm{i}}-3\hat{\mathrm{j}}$ and $\overrightarrow{B}=8\hat{\mathrm{i}}+8\hat{\mathrm{j}}$ will be

The three vectors $\overrightarrow{A}=3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{B}=\hat{\mathrm{i}}-3\hat{\mathrm{j}}+5\hat{\mathrm{k}}$ and $\overrightarrow{C}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-4\hat{\mathrm{k}}$ form

A person moves $30 \mathrm{m}$ north and then $20 \mathrm{m}$ towards east and finally $30\sqrt{2} \mathrm{m}$ in south-west direction. The displacement of the person from the origin will be