If `|veca|=sqrt(26), |vecb|=7 and |vecaxxvecb|=35, find veca.vecb`

Answer

To solve the problem, we need to find the dot product of two vectors \(\vec{a}\) and \(\vec{b}\) given the magnitudes of the vectors and the magnitude of their cross product.

### Given:
- \(|\vec{a}| = \sqrt{26}\)
- \(|\vec{b}| = 7\)
- \(|\vec{a} \times \vec{b}| = 35\)

### Step 1: Use the formula for the magnitude of the cross product
The magnitude of the cross product of two vectors is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
where \(\theta\) is the angle between the two vectors.

### Step 2: Substitute the known values
Substituting the known values into the formula:
\[ 35 = |\vec{a}| |\vec{b}| \sin \theta \]
\[ 35 = \sqrt{26} \cdot 7 \cdot \sin \theta \]

### Step 3: Solve for \(\sin \theta\)
Now, we can isolate \(\sin \theta\):
\[ \sin \theta = \frac{35}{\sqrt{26} \cdot 7} \]
\[ \sin \theta = \frac{35}{7\sqrt{26}} = \frac{5}{\sqrt{26}} \]

### Step 4: Use the Pythagorean identity to find \(\cos \theta\)
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Calculating \(\sin^2 \theta\):
\[ \sin^2 \theta = \left(\frac{5}{\sqrt{26}}\right)^2 = \frac{25}{26} \]
Now substituting into the Pythagorean identity:
\[ \frac{25}{26} + \cos^2 \theta = 1 \]
\[ \cos^2 \theta = 1 - \frac{25}{26} = \frac{1}{26} \]
Taking the square root:
\[ \cos \theta = \frac{1}{\sqrt{26}} \]

### Step 5: Calculate the dot product \(\vec{a} \cdot \vec{b}\)
The dot product is given by:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Substituting the known values:
\[ \vec{a} \cdot \vec{b} = \sqrt{26} \cdot 7 \cdot \frac{1}{\sqrt{26}} \]
\[ \vec{a} \cdot \vec{b} = 7 \]

### Final Answer:
\[ \vec{a} \cdot \vec{b} = 7 \]