The direction cosines of a vector `vecr` which is equally inclined with OX,OY and OZ are

Options

A. `+-1/(sqrt(3)),+-1/(sqrt(3)),+-1/(sqrt(3))`
B. `+-1/3,+-1/3,+-1/3`
C. `+-1/(sqrt(2)),+-1/2,+-1/2`
D. none of these

Explanation

To find the direction cosines of a vector \( \vec{r} \) that is equally inclined with the OX, OY, and OZ axes, we can follow these steps:

### Step 1: Understand the Concept of Direction Cosines
Direction cosines are the cosines of the angles that a vector makes with the coordinate axes. If a vector is equally inclined with all three axes, then the angles it makes with each axis are the same.

### Step 2: Set Up the Equations
Let the direction cosines of the vector \( \vec{r} \) be \( L, M, N \). Since the vector is equally inclined to the three axes, we have:
\[ L = M = N \]

### Step 3: Use the Property of Direction Cosines
We know that the sum of the squares of the direction cosines equals 1:
\[ L^2 + M^2 + N^2 = 1 \]
Substituting \( L = M = N \) into the equation gives:
\[ L^2 + L^2 + L^2 = 1 \]
This simplifies to:
\[ 3L^2 = 1 \]

### Step 4: Solve for L
Now, we can solve for \( L^2 \):
\[ L^2 = \frac{1}{3} \]
Taking the square root of both sides, we find:
\[ L = \pm \frac{1}{\sqrt{3}} \]

### Step 5: Determine the Direction Cosines
Since \( L = M = N \), we can conclude that:
\[ L = M = N = \pm \frac{1}{\sqrt{3}} \]
Thus, the direction cosines of the vector \( \vec{r} \) are:
\[ \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \quad \text{or} \quad \left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) \]

### Final Answer
The direction cosines of the vector \( \vec{r} \) which is equally inclined with OX, OY, and OZ axes are:
\[ \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \quad \text{and} \quad \left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) \]

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