Assertion: `|veca+vecb|lt|veca-vecb|`, Reason: `|veca+vecb|^2=|veca|^2+|vecb|^2+2veca.vecb.` (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

Answer

To solve the problem, we need to analyze the assertion and the reason given in the question step by step.

### Step 1: Understand the Assertion
The assertion states that:
\[ |\vec{a} + \vec{b}| < |\vec{a} - \vec{b}| \]

This means that the magnitude of the vector sum of \(\vec{a}\) and \(\vec{b}\) is less than the magnitude of the vector difference of \(\vec{a}\) and \(\vec{b}\).

### Step 2: Square Both Sides of the Assertion
To analyze the assertion, we can square both sides:
\[ |\vec{a} + \vec{b}|^2 < |\vec{a} - \vec{b}|^2 \]

### Step 3: Use the Formula for Magnitudes
Using the formula for the magnitude of a vector, we have:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} \]
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} \]

### Step 4: Substitute into the Inequality
Substituting these expressions into our inequality gives:
\[ |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} < |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} \]

### Step 5: Simplify the Inequality
Now, we can simplify this inequality:
- Cancel \( |\vec{a}|^2 + |\vec{b}|^2 \) from both sides:
\[ 2\vec{a} \cdot \vec{b} < -2\vec{a} \cdot \vec{b} \]
- This simplifies to:
\[ 4\vec{a} \cdot \vec{b} < 0 \]
- This means:
\[ \vec{a} \cdot \vec{b} < 0 \]

### Step 6: Conclusion about the Assertion
The assertion \( |\vec{a} + \vec{b}| < |\vec{a} - \vec{b}| \) is true only if the dot product \( \vec{a} \cdot \vec{b} \) is negative, which implies that \(\vec{a}\) and \(\vec{b}\) are in opposite directions. However, we cannot conclude that this is always true for any vectors \(\vec{a}\) and \(\vec{b}\). Thus, the assertion is **not universally true**.

### Step 7: Analyze the Reason
The reason states:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} \]
This statement is indeed true as it is a standard result from vector algebra.

### Final Conclusion
- The assertion is **false** because it does not hold true for all vectors.
- The reason is **true** as it correctly states the formula for the magnitude of the sum of two vectors.

Thus, the correct answer is:
**(D) A is false but R is true.**