The position vectors of A and B are `3hati - hatj +7hatk and 4hati-3hatj-hatk`. Find the magnitude and direction cosines of `vec(AB)`.

Answer

To solve the problem, we need to find the position vector of vector AB, its magnitude, and the direction cosines.

### Step 1: Find the position vectors of A and B
The position vectors are given as:
- Position vector of A: \(\vec{A} = 3\hat{i} - \hat{j} + 7\hat{k}\)
- Position vector of B: \(\vec{B} = 4\hat{i} - 3\hat{j} - \hat{k}\)

### Step 2: Calculate the position vector of AB
The position vector of AB can be calculated using the formula:
\[ \vec{AB} = \vec{B} - \vec{A} \]
Substituting the values:
\[ \vec{AB} = (4\hat{i} - 3\hat{j} - \hat{k}) - (3\hat{i} - \hat{j} + 7\hat{k}) \]
Now, simplify this:
\[ \vec{AB} = (4 - 3)\hat{i} + (-3 + 1)\hat{j} + (-1 - 7)\hat{k} \]
\[ \vec{AB} = 1\hat{i} - 2\hat{j} - 8\hat{k} \]

### Step 3: Find the magnitude of vector AB
The magnitude of a vector \(\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}\) is given by:
\[ |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \]
For \(\vec{AB} = 1\hat{i} - 2\hat{j} - 8\hat{k}\):
\[ |\vec{AB}| = \sqrt{1^2 + (-2)^2 + (-8)^2} \]
Calculating this:
\[ |\vec{AB}| = \sqrt{1 + 4 + 64} = \sqrt{69} \]

### Step 4: Find the direction cosines of vector AB
The direction cosines are given by:
\[ \cos \alpha = \frac{a}{|\vec{AB}|}, \quad \cos \beta = \frac{b}{|\vec{AB}|}, \quad \cos \gamma = \frac{c}{|\vec{AB}|} \]
where \(a\), \(b\), and \(c\) are the components of vector \(\vec{AB}\).

For \(\vec{AB} = 1\hat{i} - 2\hat{j} - 8\hat{k}\):
- \(a = 1\)
- \(b = -2\)
- \(c = -8\)

Now substituting into the formulas:
\[ \cos \alpha = \frac{1}{\sqrt{69}}, \quad \cos \beta = \frac{-2}{\sqrt{69}}, \quad \cos \gamma = \frac{-8}{\sqrt{69}} \]

### Final Result
- The magnitude of \(\vec{AB}\) is \(\sqrt{69}\).
- The direction cosines are:
- \(\cos \alpha = \frac{1}{\sqrt{69}}\)
- \(\cos \beta = \frac{-2}{\sqrt{69}}\)
- \(\cos \gamma = \frac{-8}{\sqrt{69}}\)