Find the cosines of the angles made by `bara = 2bari+3barj+6bark` with the X, Y, Z - axes.

Answer

To find the cosines of the angles made by the vector \(\vec{a} = 2\hat{i} + 3\hat{j} + 6\hat{k}\) with the X, Y, and Z axes, we can follow these steps:

### Step 1: Calculate the modulus of the vector \(\vec{a}\)

The modulus of a vector \(\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\) is given by:

\[ |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \]

For our vector:

\[ |\vec{a}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Step 2: Find the cosine of the angle with the X-axis

The cosine of the angle \(\theta_1\) made with the X-axis is given by the formula:

\[ \cos \theta_1 = \frac{\vec{a} \cdot \hat{i}}{|\vec{a}| \cdot |\hat{i}|} \]

Here, \(\hat{i}\) is the unit vector along the X-axis, and its modulus is 1. The dot product \(\vec{a} \cdot \hat{i}\) is:

\[ \vec{a} \cdot \hat{i} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot \hat{i} = 2 \]

Thus,

\[ \cos \theta_1 = \frac{2}{7} \]

### Step 3: Find the cosine of the angle with the Y-axis

Similarly, for the Y-axis, we have:

\[ \cos \theta_2 = \frac{\vec{a} \cdot \hat{j}}{|\vec{a}| \cdot |\hat{j}|} \]

The dot product \(\vec{a} \cdot \hat{j}\) is:

\[ \vec{a} \cdot \hat{j} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot \hat{j} = 3 \]

Thus,

\[ \cos \theta_2 = \frac{3}{7} \]

### Step 4: Find the cosine of the angle with the Z-axis

For the Z-axis, we have:

\[ \cos \theta_3 = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}| \cdot |\hat{k}|} \]

The dot product \(\vec{a} \cdot \hat{k}\) is:

\[ \vec{a} \cdot \hat{k} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot \hat{k} = 6 \]

Thus,

\[ \cos \theta_3 = \frac{6}{7} \]

### Final Result

The cosines of the angles made by the vector \(\vec{a}\) with the X, Y, and Z axes are:

\[ \cos \theta_1 = \frac{2}{7}, \quad \cos \theta_2 = \frac{3}{7}, \quad \cos \theta_3 = \frac{6}{7} \]