If `vec(OP)=2hati+3hatj-hatk and vec(OQ)=3hati-4hatj+2hatk` find the modulus and direction cosines of `vec(PQ)`.

Answer

To solve the problem of finding the modulus and direction cosines of the vector \(\vec{PQ}\) given the vectors \(\vec{OP}\) and \(\vec{OQ}\), we will follow these steps:

### Step 1: Determine the vectors \(\vec{OP}\) and \(\vec{OQ}\)

Given:
\[ \vec{OP} = 2\hat{i} + 3\hat{j} - \hat{k} \]
\[ \vec{OQ} = 3\hat{i} - 4\hat{j} + 2\hat{k} \]

### Step 2: Calculate the vector \(\vec{PQ}\)

The vector \(\vec{PQ}\) can be found using the formula:
\[ \vec{PQ} = \vec{OQ} - \vec{OP} \]

Substituting the values:
\[ \vec{PQ} = (3\hat{i} - 4\hat{j} + 2\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k}) \]

### Step 3: Simplify the expression for \(\vec{PQ}\)

Now, we perform the subtraction component-wise:
\[ \vec{PQ} = (3 - 2)\hat{i} + (-4 - 3)\hat{j} + (2 + 1)\hat{k} \]
\[ \vec{PQ} = 1\hat{i} - 7\hat{j} + 3\hat{k} \]

### Step 4: Find the modulus of \(\vec{PQ}\)

The modulus of a vector \(\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}\) is given by:
\[ |\vec{A}| = \sqrt{a^2 + b^2 + c^2} \]

For \(\vec{PQ} = 1\hat{i} - 7\hat{j} + 3\hat{k}\):
\[ |\vec{PQ}| = \sqrt{1^2 + (-7)^2 + 3^2} \]
\[ |\vec{PQ}| = \sqrt{1 + 49 + 9} = \sqrt{59} \]

### Step 5: Calculate the direction cosines of \(\vec{PQ}\)

The direction cosines are given by:
\[ \cos \alpha = \frac{a}{|\vec{A}|}, \quad \cos \beta = \frac{b}{|\vec{A}|}, \quad \cos \gamma = \frac{c}{|\vec{A}|} \]

For \(\vec{PQ} = 1\hat{i} - 7\hat{j} + 3\hat{k}\):
- \(a = 1\)
- \(b = -7\)
- \(c = 3\)

Thus, the direction cosines are:
\[ \cos \alpha = \frac{1}{\sqrt{59}}, \quad \cos \beta = \frac{-7}{\sqrt{59}}, \quad \cos \gamma = \frac{3}{\sqrt{59}} \]

### Final Result

- The modulus of \(\vec{PQ}\) is \(\sqrt{59}\).
- The direction cosines of \(\vec{PQ}\) are:
- \(\cos \alpha = \frac{1}{\sqrt{59}}\)
- \(\cos \beta = \frac{-7}{\sqrt{59}}\)
- \(\cos \gamma = \frac{3}{\sqrt{59}}\)