What is the point of trisection of PQ nearer to P if positions of P and Q are `3 hati + 3 hatj- 4 hatk and 9 hati + 8hatj-10 hatk` respectively ?

Answer

To find the point of trisection of the line segment PQ that is closer to point P, we will follow these steps:

1. **Identify the Position Vectors**:
The position vectors of points P and Q are given as:
- \( \vec{P} = 3 \hat{i} + 3 \hat{j} - 4 \hat{k} \)
- \( \vec{Q} = 9 \hat{i} + 8 \hat{j} - 10 \hat{k} \)

2. **Understand the Trisection**:
A trisection means dividing the line segment into three equal parts. The point nearer to P will divide the segment in the ratio 1:2.

3. **Apply the Section Formula**:
The section formula states that if a point divides a line segment joining two points in the ratio \( m:n \), then the position vector of the point is given by:
\[ \vec{R} = \frac{n \vec{P} + m \vec{Q}}{m+n} \]
For our case, \( m = 2 \) and \( n = 1 \) (since we want the point nearer to P).

4. **Substitute the Values**:
Substitute the values into the section formula:
\[ \vec{K} = \frac{1 \cdot \vec{P} + 2 \cdot \vec{Q}}{1 + 2} \]
\[ = \frac{1 \cdot (3 \hat{i} + 3 \hat{j} - 4 \hat{k}) + 2 \cdot (9 \hat{i} + 8 \hat{j} - 10 \hat{k})}{3} \]

5. **Calculate Each Component**:
- **For \( \hat{i} \)**:
\[ 1 \cdot 3 + 2 \cdot 9 = 3 + 18 = 21 \]
- **For \( \hat{j} \)**:
\[ 1 \cdot 3 + 2 \cdot 8 = 3 + 16 = 19 \]
- **For \( \hat{k} \)**:
\[ 1 \cdot (-4) + 2 \cdot (-10) = -4 - 20 = -24 \]

6. **Combine the Results**:
Now, substitute these back into the equation:
\[ \vec{K} = \frac{21 \hat{i} + 19 \hat{j} - 24 \hat{k}}{3} \]

7. **Final Calculation**:
Divide each component by 3:
\[ \vec{K} = 7 \hat{i} + \frac{19}{3} \hat{j} - 8 \hat{k} \]

Thus, the position vector of the point of trisection nearer to P is:
\[ \vec{K} = 7 \hat{i} + \frac{19}{3} \hat{j} - 8 \hat{k} \]