If the position vector of one end of the line segment AB be `2hati+3hatj-hatk` and the position vector of its middle point be `3(hati+hatj+hatk)`, then find the position vector of the other end.

Answer

To find the position vector of the other end of the line segment AB, we can follow these steps:

### Step 1: Define the given vectors
Let the position vector of point A (one end of the segment) be:
\[ \vec{A} = 2\hat{i} + 3\hat{j} - \hat{k} \]
Let the position vector of point B (the other end of the segment) be:
\[ \vec{B} = x\hat{i} + y\hat{j} + z\hat{k} \]
The position vector of the midpoint R is given as:
\[ \vec{R} = 3(\hat{i} + \hat{j} + \hat{k}) = 3\hat{i} + 3\hat{j} + 3\hat{k} \]

### Step 2: Use the midpoint formula
The midpoint R of the line segment AB can be calculated using the formula:
\[ \vec{R} = \frac{\vec{A} + \vec{B}}{2} \]
Substituting the known values:
\[ 3\hat{i} + 3\hat{j} + 3\hat{k} = \frac{(2\hat{i} + 3\hat{j} - \hat{k}) + (x\hat{i} + y\hat{j} + z\hat{k})}{2} \]

### Step 3: Multiply both sides by 2
To eliminate the fraction, multiply both sides by 2:
\[ 2(3\hat{i} + 3\hat{j} + 3\hat{k}) = (2\hat{i} + 3\hat{j} - \hat{k}) + (x\hat{i} + y\hat{j} + z\hat{k}) \]
This simplifies to:
\[ 6\hat{i} + 6\hat{j} + 6\hat{k} = (2 + x)\hat{i} + (3 + y)\hat{j} + (-1 + z)\hat{k} \]

### Step 4: Equate coefficients
Now, we can equate the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\):
1. For \(\hat{i}\):
\[ 2 + x = 6 \implies x = 6 - 2 = 4 \]
2. For \(\hat{j}\):
\[ 3 + y = 6 \implies y = 6 - 3 = 3 \]
3. For \(\hat{k}\):
\[ -1 + z = 6 \implies z = 6 + 1 = 7 \]

### Step 5: Write the position vector of point B
Now substituting the values of \(x\), \(y\), and \(z\) back into the vector \(\vec{B}\):
\[ \vec{B} = 4\hat{i} + 3\hat{j} + 7\hat{k} \]

### Final Answer
The position vector of the other end of the line segment AB is:
\[ \vec{B} = 4\hat{i} + 3\hat{j} + 7\hat{k} \]

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