Let a,b,c be three distinct positive real numbers. If `vecp,vecq,vecr` lie in a plane, where `vecp=ahati-ahatj+bhatk, vecq=hati+hatk` and `vecr=chati+chatj+bhatk`, then b is

Options

A. The A.M. of a,c
B. The G.M. of a, c
C. The H.M. of a, c
D. Equal to 0

Answer

To solve the problem, we need to determine the value of \( b \) given that the vectors \( \vec{p}, \vec{q}, \vec{r} \) lie in the same plane. This means that the scalar triple product of these vectors must be zero.

### Step-by-step Solution:

1. **Define the Vectors**:
\[ \vec{p} = a \hat{i} - a \hat{j} + b \hat{k} \]
\[ \vec{q} = \hat{i} + \hat{k} \]
\[ \vec{r} = c \hat{i} + c \hat{j} + b \hat{k} \]

2. **Set Up the Scalar Triple Product**:
The scalar triple product of vectors \( \vec{p}, \vec{q}, \vec{r} \) can be expressed using the determinant of a matrix formed by the components of these vectors:
\[ \begin{vmatrix} a & -a & b \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
We will expand the determinant:
\[ = a \begin{vmatrix} 0 & 1 \\ c & b \end{vmatrix} - (-a) \begin{vmatrix} 1 & 1 \\ c & b \end{vmatrix} + b \begin{vmatrix} 1 & 0 \\ c & c \end{vmatrix} \]
\[ = a(0 \cdot b - 1 \cdot c) + a(1 \cdot b - 1 \cdot c) + b(1 \cdot c - 0 \cdot c) \]
\[ = -ac + a(b - c) + bc \]
\[ = ab - ac + bc - ac = ab + bc - 2ac \]

4. **Set the Determinant to Zero**:
Since the vectors are coplanar:
\[ ab + bc - 2ac = 0 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ ab + bc = 2ac \]
\[ b(a + c) = 2ac \]
\[ b = \frac{2ac}{a + c} \]

6. **Identifying the Mean**:
The expression \( \frac{2ac}{a + c} \) is the formula for the harmonic mean of \( a \) and \( c \).

### Conclusion:
Thus, we conclude that \( b \) is the harmonic mean of \( a \) and \( c \).