If the position vectors of A and B are `hati+3hatj-7hatk and 5hati-2hatj+4hatk`, then the direction cosine of AB along Y-axis is

Options

A. `(4)/(sqrt(162))`
B. `-(5)/(sqrt(162))`
C. `-5`
D. 11

Explanation

To find the direction cosine of the vector AB along the Y-axis, we will follow these steps:

### Step 1: Identify the position vectors of points A and B.
The position vector of point A is given as:
\[ \vec{A} = \hat{i} + 3\hat{j} - 7\hat{k} \]
The position vector of point B is given as:
\[ \vec{B} = 5\hat{i} - 2\hat{j} + 4\hat{k} \]

### Step 2: Calculate the vector AB.
The vector AB can be calculated using the formula:
\[ \vec{AB} = \vec{B} - \vec{A} \]
Substituting the position vectors:
\[ \vec{AB} = (5\hat{i} - 2\hat{j} + 4\hat{k}) - (\hat{i} + 3\hat{j} - 7\hat{k}) \]
This simplifies to:
\[ \vec{AB} = (5 - 1)\hat{i} + (-2 - 3)\hat{j} + (4 + 7)\hat{k} \]
\[ \vec{AB} = 4\hat{i} - 5\hat{j} + 11\hat{k} \]

### Step 3: Calculate the magnitude of vector AB.
The magnitude of vector AB is given by:
\[ |\vec{AB}| = \sqrt{(4)^2 + (-5)^2 + (11)^2} \]
Calculating this:
\[ |\vec{AB}| = \sqrt{16 + 25 + 121} = \sqrt{162} \]

### Step 4: Find the unit vector of AB.
The unit vector \(\hat{u}_{AB}\) in the direction of AB is given by:
\[ \hat{u}_{AB} = \frac{\vec{AB}}{|\vec{AB}|} \]
Substituting the values:
\[ \hat{u}_{AB} = \frac{4\hat{i} - 5\hat{j} + 11\hat{k}}{\sqrt{162}} \]
This can be expressed as:
\[ \hat{u}_{AB} = \frac{4}{\sqrt{162}}\hat{i} - \frac{5}{\sqrt{162}}\hat{j} + \frac{11}{\sqrt{162}}\hat{k} \]

### Step 5: Identify the direction cosine along the Y-axis.
The direction cosines are represented as \(\cos \alpha\), \(\cos \beta\), and \(\cos \gamma\) for the x, y, and z axes respectively. Here, we need \(\cos \beta\), which corresponds to the Y-axis:
\[ \cos \beta = -\frac{5}{\sqrt{162}} \]

### Final Answer:
Thus, the direction cosine of AB along the Y-axis is:
\[ \cos \beta = -\frac{5}{\sqrt{162}} \]

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