Q60 / 155 · WBJEE 2026 · Mathematics
The position vectors of two adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OB}$ 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 $\tan\left(\dfrac{\theta}{2}\right)$ is
- Correct answerA$\dfrac{1}{3}$
- B$\dfrac{1}{\sqrt{3}}$
- C$\sqrt{3}$
- D$1$
Solution
Let $|\vec{a}| = x$ and $|\vec{b}| = y$. Since OACB is a rectangle, the adjacent sides $\vec{a}$ and $\vec{b}$ are perpendicular. Thus, $\vec{a} \cdot \vec{b} = 0$ and $|\vec{a} \times \vec{b}| = xy$.
Given $16|\vec{a} \times \vec{b}| = 3(|\vec{a}| + |\vec{b}|)^2$, we substitute the magnitudes:
$16xy = 3(x+y)^2$
$16xy = 3x^2 + 6xy + 3y^2$
$3x^2 - 10xy + 3y^2 = 0$
$(3x - y)(x - 3y) = 0 \Rightarrow y = 3x$ or $x = 3y$.
The diagonals of the rectangle are given by the vectors $\overrightarrow{OC} = \vec{a} + \vec{b}$ and $\overrightarrow{AB} = \vec{b} - \vec{a}$.
Let $\theta$ be the acute angle between the diagonals. The cosine of this angle is:
$\cos \theta = \left| \dfrac{\overrightarrow{OC} \cdot \overrightarrow{AB}}{|\overrightarrow{OC}| |\overrightarrow{AB}|} \right| = \left| \dfrac{(\vec{b} + \vec{a}) \cdot (\vec{b} - \vec{a})}{|\vec{b} + \vec{a}| |\vec{b} - \vec{a}|} \right|$
$\cos \theta = \left| \dfrac{|\vec{b}|^2 - |\vec{a}|^2}{\sqrt{|\vec{b}|^2 + |\vec{a}|^2} \sqrt{|\vec{b}|^2 + |\vec{a}|^2}} \right| = \left| \dfrac{y^2 - x^2}{y^2 + x^2} \right|$
Substituting $y = 3x$ (the result is the same for $x = 3y$):
$\cos \theta = \left| \dfrac{9x^2 - x^2}{9x^2 + x^2} \right| = \dfrac{8x^2}{10x^2} = \dfrac{4}{5}$
Using the half-angle formula for tangent:
$\tan\left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}}$
$\tan\left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1 - 4/5}{1 + 4/5}} = \sqrt{\dfrac{1/5}{9/5}} = \sqrt{\dfrac{1}{9}} = \dfrac{1}{3}$
Answer: $\dfrac{1}{3}$
Given $16|\vec{a} \times \vec{b}| = 3(|\vec{a}| + |\vec{b}|)^2$, we substitute the magnitudes:
$16xy = 3(x+y)^2$
$16xy = 3x^2 + 6xy + 3y^2$
$3x^2 - 10xy + 3y^2 = 0$
$(3x - y)(x - 3y) = 0 \Rightarrow y = 3x$ or $x = 3y$.
The diagonals of the rectangle are given by the vectors $\overrightarrow{OC} = \vec{a} + \vec{b}$ and $\overrightarrow{AB} = \vec{b} - \vec{a}$.
Let $\theta$ be the acute angle between the diagonals. The cosine of this angle is:
$\cos \theta = \left| \dfrac{\overrightarrow{OC} \cdot \overrightarrow{AB}}{|\overrightarrow{OC}| |\overrightarrow{AB}|} \right| = \left| \dfrac{(\vec{b} + \vec{a}) \cdot (\vec{b} - \vec{a})}{|\vec{b} + \vec{a}| |\vec{b} - \vec{a}|} \right|$
$\cos \theta = \left| \dfrac{|\vec{b}|^2 - |\vec{a}|^2}{\sqrt{|\vec{b}|^2 + |\vec{a}|^2} \sqrt{|\vec{b}|^2 + |\vec{a}|^2}} \right| = \left| \dfrac{y^2 - x^2}{y^2 + x^2} \right|$
Substituting $y = 3x$ (the result is the same for $x = 3y$):
$\cos \theta = \left| \dfrac{9x^2 - x^2}{9x^2 + x^2} \right| = \dfrac{8x^2}{10x^2} = \dfrac{4}{5}$
Using the half-angle formula for tangent:
$\tan\left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}}$
$\tan\left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1 - 4/5}{1 + 4/5}} = \sqrt{\dfrac{1/5}{9/5}} = \sqrt{\dfrac{1}{9}} = \dfrac{1}{3}$
Answer: $\dfrac{1}{3}$
Hence, the correct answer is option: 'A'.