if the vectors \(\underline{a}\) and \(\underline{b}\) such that \(|\underline{\mathrm{a}}+\underline{\mathrm{b}}|<|\underline{\mathrm{a}}-\underline{\mathrm{b}}|\) the angle between \(\underline{a}\) and \(\underline{b}\) is. (a) acute (b) right angle (c) obtuse (d) none of these.

We will square both sides of the given inequality to remove the absolute value sign: \( ( | \underline{a} + \underline{b} | )^2 < ( | \underline{a} - \underline{b} | )^2 \). Now, apply the triangle inequality theorem, which states that the square of the magnitude of the sum of two vectors equals the sum of the squares of their magnitudes plus two times the product of their magnitudes and the cosine of the angle between them: \( | \underline{a} + \underline{b} |^2 = |\underline{a}|^2 + |\underline{b}|^2 + 2 |\underline{a}| |\underline{b}| \cos\theta \). Similarly, for \( | \underline{a} - \underline{b} |^2 \), we have \( | \underline{a} - \underline{b} |^2 = |\underline{a}|^2 + |\underline{b}|^2 - 2 |\underline{a}| |\underline{b}| \cos\theta \).

Substitute the expressions obtained in step 1 into the inequality: \( |\underline{a}|^2 + |\underline{b}|^2 + 2 |\underline{a}| |\underline{b}| \cos\theta < |\underline{a}|^2 + |\underline{b}|^2 - 2 |\underline{a}| |\underline{b}| \cos\theta \).

Now, cancel the common terms on both sides: \( 2 |\underline{a}| |\underline{b}| \cos\theta < - 2 |\underline{a}| |\underline{b}| \cos\theta \). Divide both sides by \( 2 |\underline{a}| |\underline{b}| \): \( \cos\theta < - \cos\theta \).

From \( \cos\theta < - \cos\theta \), we can conclude that the angle θ between vectors \( \underline{a} \) and \( \underline{b} \) must be obtuse since the cosine of an obtuse angle is negative, and the inequality states that \( \cos\theta \) must be less than a negative value. Therefore, the correct answer is: (c) obtuse