Question

There are two vectors $\overrightarrow{A}=3\hat{i}+\hat{j}$ and $\overrightarrow{B}=\hat{j}+2\hat{k}$. For these two vectors
(i) find the component of $\overrightarrow{A}$ along $\overrightarrow{B}$ and perpendicular to $\overrightarrow{B}$ in vector form.
(ii) If $\overrightarrow{A}$ and $\overrightarrow{B}$ are the adjacent sides of a parallelogram then find the magnitude of its area.
(iii) find a unit vector which is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$.

• A. $\frac{1}{5}\left(\begin{array}{c}\hat{j}+2\hat{k}\end{array}\right),3\hat{i}+\frac{4}{5}\hat{j}-\frac{2}{5}\hat{k}$ ; $7units$ ;$\frac{2}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{3}{7}\hat{k}$
• B. $\frac{1}{4}\left(\begin{array}{c}\hat{j}+2\hat{k}\end{array}\right),3\hat{i}+\frac{4}{5}\hat{j}-\frac{2}{5}\hat{k}$ ; $9units$ ;$\frac{2}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{3}{7}\hat{k}$
• C. $\frac{1}{5}\left(\begin{array}{c}\hat{j}+2\hat{k}\end{array}\right),\hat{i}+\frac{4}{5}\hat{j}-\frac{2}{5}\hat{k}$ ; $9units$ ;$\frac{2}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{3}{7}\hat{k}$
• D. $\frac{1}{4}\left(\begin{array}{c}\hat{j}+2\hat{k}\end{array}\right),3\hat{i}+\frac{4}{5}\hat{j}-\frac{2}{5}\hat{k}$ ; $7units$ ;$\frac{3}{7}\hat{i}-\frac{5}{7}\hat{j}+\frac{3}{7}\hat{k}$

Answer 1

The correct option is A $\frac{1}{5}\left(\begin{array}{c}\hat{j}+2\hat{k}\end{array}\right),3\hat{i}+\frac{4}{5}\hat{j}-\frac{2}{5}\hat{k}$ ; $7units$ ;$\frac{2}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{3}{7}\hat{k}$

The component of $\overrightarrow{A}$ along $\overrightarrow{B}$

$=\frac{\overrightarrow{A}.\overrightarrow{B}}{\overrightarrow{B}.\overrightarrow{B}}\overrightarrow{B}$

$=\dfrac{1}{{1+4}}(\hat j+2\hat k=\dfrac{1}{5}(\hat j+2\hat k)$

The component of $\overrightarrow{A}$ along perpendicular to $\overrightarrow{B}=\overrightarrow{A}-\frac{\overrightarrow{A}.\overrightarrow{B}}{\overrightarrow{B}.\overrightarrow{B}}\overrightarrow{B}=(3\hat{i}+\hat{j})-\frac{1}{5}(\hat{j}+2\hat{k})=3\hat{i}+\frac{4}{5}\hat{j}-\frac{2}{5}\hat{k}$

The magnitude of the area $=|\overrightarrow{A}\times \overrightarrow{B}|=|2\hat{i}-6\hat{j}+3\hat{k}|=\sqrt{2^2+(-6{)}^2+3^2}=7$ units

The vector from cross product of A and B is perpendicular to both A and B vectors.

Thus, unit vector, $\hat{n}=\frac{\overrightarrow{A}\times \overrightarrow{B}}{|\overrightarrow{A}\times \overrightarrow{B}|}=\frac{2\hat{i}-6\hat{j}+3\hat{k}}{7}$