A point `(alpha, beta, gamma)` satisfies the equation of the plane `3x+4y+7z=3.` The value of `beta`, such that `vecp=alphahati+betahatj+gammahatk` satisfies the relation `hatjxx(hatjxxvecp)=vec0`, is equal to

Answer

To solve the problem step by step, we need to find the value of \( \beta \) such that the point \( (\alpha, \beta, \gamma) \) satisfies the equation of the plane \( 3x + 4y + 7z = 3 \) and the vector \( \vec{p} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \) satisfies the relation \( \hat{j} \times (\hat{j} \times \vec{p}) = \vec{0} \).

### Step 1: Use the plane equation

The point \( (\alpha, \beta, \gamma) \) satisfies the plane equation:

\[ 3\alpha + 4\beta + 7\gamma = 3 \quad \text{(Equation 1)} \]

### Step 2: Analyze the triple product condition

The condition \( \hat{j} \times (\hat{j} \times \vec{p}) = \vec{0} \) implies that the vector \( \vec{p} \) must be parallel to \( \hat{j} \). This means that the \( \hat{i} \) and \( \hat{k} \) components of \( \vec{p} \) must be zero.

### Step 3: Calculate \( \hat{j} \cdot \vec{p} \)

The vector \( \vec{p} \) can be expressed as:

\[ \vec{p} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \]

Now, we calculate the dot product \( \hat{j} \cdot \vec{p} \):

\[ \hat{j} \cdot \vec{p} = \hat{j} \cdot (\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}) = \beta \]

### Step 4: Calculate \( \hat{j} \cdot \hat{j} \)

We know that:

\[ \hat{j} \cdot \hat{j} = 1 \]

### Step 5: Substitute into the triple product formula

Using the property of the triple product, we have:

\[ \hat{j} \times (\hat{j} \times \vec{p}) = (\hat{j} \cdot \vec{p}) \hat{j} - (\hat{j} \cdot \hat{j}) \vec{p} \]

Substituting the values we calculated:

\[ \hat{j} \times (\hat{j} \times \vec{p}) = \beta \hat{j} - \vec{p} \]

Setting this equal to \( \vec{0} \):

\[ \beta \hat{j} - \vec{p} = \vec{0} \]

This implies:

\[ \vec{p} = \beta \hat{j} \]

### Step 6: Equate components of \( \vec{p} \)

From \( \vec{p} = \beta \hat{j} \), we can equate the components:

- The \( \hat{i} \) component gives \( \alpha = 0 \)
- The \( \hat{j} \) component gives \( \beta = \beta \) (which is consistent)
- The \( \hat{k} \) component gives \( \gamma = 0 \)

### Step 7: Substitute back into the plane equation

Substituting \( \alpha = 0 \) and \( \gamma = 0 \) into Equation 1:

\[ 3(0) + 4\beta + 7(0) = 3 \]

This simplifies to:

\[ 4\beta = 3 \]

### Step 8: Solve for \( \beta \)

Dividing both sides by 4:

\[ \beta = \frac{3}{4} = 0.75 \]

Thus, the required value of \( \beta \) is:

\[ \boxed{0.75} \]