Question

The position vectors of two adjacent sides of a rectangle \(OACB\) are \(\vec{a}\) and \(\vec{b}\) respectively, where \(O\) is the origin. If \(16|\vec{a}\times\vec{b}|=3(|\vec{a}|+|\vec{b}|)^{2}\) and \(\theta\) be the acute angle between the diagonals \(OC\) and \(AB\), then the value of \(\tan\left(\frac{\theta}{2}\right)\) is:

• A. $\frac{1}{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. 1

Hint:

In any rectangle where the adjacent side vectors form a length ratio $\frac{b}{a} = k$, the acute angle $\theta$ between its diagonals can be found directly using the handy shortcut formula: $\tan\left(\frac{\theta}{2}\right) = \frac{1}{k}$ (assuming $k > 1$). This saves you from performing long vector dot-product expansions midway through the exam!

Answer 1

The Correct Option is A

Concept: In a rectangle $OACB$, the adjacent side vectors $\vec{a}$ and $\vec{b}$ are perpendicular ($\vec{a} \cdot \vec{b} = 0$), which means the angle between them is $90^\circ$. We can find the relationship between their lengths by expanding the given vector cross-product equation, and then use trigonometric identities to find the angle between the diagonals. Step 1: Expand the vector equation using side lengths.
Let the lengths of the sides be defined as $a = |\vec{a}|$ and $b = |\vec{b}|$. Since the sides meet at a right angle, the magnitude of their cross product simplifies to: $$|\vec{a} \times \vec{b}| = a \cdot b \cdot \sin(90^\circ) = ab$$ Substitute this back into the problem's core equation: $$16ab = 3(a + b)^2 \quad \Rightarrow \quad 16ab = 3a^2 + 6ab + 3b^2$$ Move all terms to one side to form a quadratic equation: $$3a^2 - 10ab + 3b^2 = 0$$

Step 2: Factor the equation to find the ratio between the sides.
Split the middle term to factor the quadratic expression: $$3a^2 - 9ab - ab + 3b^2 = 0 \quad \Rightarrow \quad 3a(a - 3b) - b(a - 3b) = 0$$ $$(3a - b)(a - 3b) = 0 \quad \Rightarrow \quad b = 3a \quad \text{or} \quad a = 3b$$ By symmetry, let us choose $b = 3a$ without loss of generality. This means the rectangle is 3 times taller than it is wide.

Step 3: Relate the side ratio to the diagonal half-angle.
Let $\alpha$ be the angle that the main diagonal $OC$ makes with the base side vector $\vec{a}$. Using the right-triangle properties of the rectangle: $$\tan \alpha = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{b}{a} = \frac{3a}{a} = 3$$ By standard geometry, the diagonals of a rectangle are equal in length and bisect each other, splitting the shape into isosceles triangles. The acute angle $\theta$ between the intersecting diagonals satisfies the geometric identity: $$\frac{\theta}{2} = 90^\circ - \alpha$$

Step 4: Calculate the value of $\tan(\frac{\theta}{2})$.
Apply the trigonometric complement rule to evaluate our target expression: $$\tan\left(\frac{\theta}{2}\right) = \tan(90^\circ - \alpha) = \cot \alpha = \frac{1}{\tan \alpha}$$ Substitute our known value $\tan \alpha = 3$ into the fraction: $$\tan\left(\frac{\theta}{2}\right) = \frac{1}{3}$$ This matches option (A) perfectly.