Question

A vector \(\overrightarrow{r}\) is inclined at equal angles to the three axes. If the magnitude of \(\overrightarrow{r}\) is \(3\sqrt3\) units, then the value of \(\overrightarrow{r}\) is:

• A. \(\pm3(\hat{i}+\hat{j}+\hat{k})\)
• B. \(\pm2(\hat{i}+\hat{j}+\hat{k})\)
• C. \(\pm(\hat{i}+\hat{j}+\hat{k})\)
• D. \(\pm3(\hat{i}-\hat{j}-\hat{k})\)

Answer 1

The Correct Option is A

The vector \(\overrightarrow{r}\) is inclined at equal angles with the coordinate axes. Let the direction cosines be \(l, m,\) and \(n\) for the x, y, and z axes respectively. Since the vector is equally inclined, we have:

\[l = m = n\]

From the identity of direction cosines:

\[l^2 + m^2 + n^2 = 1\]

Substituting \(l = m = n\), this becomes:

\[3l^2 = 1\]

\[l^2 = \frac{1}{3}\]
\[l = m = n = \frac{1}{\sqrt{3}}\]

The vector \(\overrightarrow{r}\) can be expressed in terms of its components along the axes as:

\[\overrightarrow{r} = r(l\hat{i} + m\hat{j} + n\hat{k})\]

Substituting the values of \(l, m, \) and \(n\):

\[\overrightarrow{r} = r\left(\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}\right)\]

\[\overrightarrow{r} = \frac{r}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\]

The magnitude of \(\overrightarrow{r}\) is given as \(3\sqrt{3}\):

\[\left|\overrightarrow{r}\right| = \left|\frac{r}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\right| = \frac{r}{\sqrt{3}} \times \sqrt{3} = r\]

Thus, the magnitude equation becomes:

\[r = 3\sqrt{3}\]

The vector \(\overrightarrow{r}\) is:

\[\overrightarrow{r} = 3(\hat{i} + \hat{j} + \hat{k})\]

Therefore, the value of \(\overrightarrow{r}\) in the correct option is:

\(\pm3(\hat{i}+\hat{j}+\hat{k})\)