Let the position vectors of points ‘A’ and ‘B’ be `hat(i)+hat(j) + hat(k)` and `2hat(i)+hat(j)+3hat(k)`, respectively. A point ‘P’ divides the line segment AB internally in the ratio `lambda:1(lambda gt 0)`. If O is the origin and `vec(OB).vec(OP)-3|vec(OA)xx vec(OP)|^(2)=6`, then `lambda` is equal to ________

Answer

To solve the problem, we will follow these steps:

1. **Identify the position vectors**:
- Let the position vector of point A be \(\vec{OA} = \hat{i} + \hat{j} + \hat{k}\).
- Let the position vector of point B be \(\vec{OB} = 2\hat{i} + \hat{j} + 3\hat{k}\).

2. **Find the position vector of point P**:
- Point P divides the line segment AB in the ratio \(\lambda:1\). By the section formula, the position vector of point P, \(\vec{OP}\), is given by:
\[ \vec{OP} = \frac{\lambda \vec{OB} + 1 \cdot \vec{OA}}{\lambda + 1} \]
Substituting the vectors:
\[ \vec{OP} = \frac{\lambda(2\hat{i} + \hat{j} + 3\hat{k}) + (\hat{i} + \hat{j} + \hat{k})}{\lambda + 1} \]
Simplifying this:
\[ \vec{OP} = \frac{(2\lambda + 1)\hat{i} + (\lambda + 1)\hat{j} + (3\lambda + 1)\hat{k}}{\lambda + 1} \]

3. **Calculate \(\vec{OB} \cdot \vec{OP}\)**:
- We need to compute \(\vec{OB} \cdot \vec{OP}\):
\[ \vec{OB} \cdot \vec{OP} = (2\hat{i} + \hat{j} + 3\hat{k}) \cdot \left(\frac{(2\lambda + 1)\hat{i} + (\lambda + 1)\hat{j} + (3\lambda + 1)\hat{k}}{\lambda + 1}\right) \]
Calculating the dot product:
\[ = \frac{1}{\lambda + 1} \left[ 2(2\lambda + 1) + 1(\lambda + 1) + 3(3\lambda + 1) \right] \]
Simplifying:
\[ = \frac{1}{\lambda + 1} \left[ 4\lambda + 2 + \lambda + 1 + 9\lambda + 3 \right] = \frac{14\lambda + 6}{\lambda + 1} \]

4. **Calculate \(|\vec{OA} \times \vec{OP}|\)**:
- We need to find \(|\vec{OA} \times \vec{OP}|\). Using the determinant form:
\[ \vec{OA} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \vec{OP} = \begin{pmatrix} \frac{2\lambda + 1}{\lambda + 1} \\ \frac{\lambda + 1}{\lambda + 1} \\ \frac{3\lambda + 1}{\lambda + 1} \end{pmatrix} \]
The cross product can be computed using the determinant:
\[ |\vec{OA} \times \vec{OP}| = \left| \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ \frac{2\lambda + 1}{\lambda + 1} & 1 & \frac{3\lambda + 1}{\lambda + 1} \end{vmatrix} \right| \]
This determinant can be computed to find the magnitude.

5. **Set up the equation**:
- Given the equation:
\[ \vec{OB} \cdot \vec{OP} - 3|\vec{OA} \times \vec{OP}|^2 = 6 \]
Substitute the expressions for \(\vec{OB} \cdot \vec{OP}\) and \(|\vec{OA} \times \vec{OP}|\) into this equation and solve for \(\lambda\).

6. **Solve for \(\lambda\)**:
- Rearranging the equation will yield a quadratic equation in \(\lambda\). Solve this quadratic equation to find the value of \(\lambda\).