Question

ABCD is a square whose vertices are A(0, 0), B(2, 0), C(2, 2) and D(0, 2). The square is rotated in the XY-plane through an angle 30⁰ in the anti-clockwise sense about an axis passing though A perpendicular to the XY-plane. The equation of the diagonal BD of this rotated square is-

• A. (2 –$\sqrt{3}$) x + y = 2$\sqrt{3}$ – 2
• B. (2 +$\sqrt{3}$) x + y = 2$\sqrt{3}$ – 1
• C. (2 –$\sqrt{3}$) x – y = 2$\sqrt{3}$ – 4
• D. (2 –$\sqrt{3}$) x – y = 2$\sqrt{3}$ + 1

Answer 1

Answer: (A)

(2 –$\sqrt{3}$) x + y = 2$\sqrt{3}$ – 2

Answer 2

We have, (see fig.)

B ŗ (2 cos 30⁰, 2 sin 30⁰) = ($\sqrt{3}$, 1)

D ŗ (–2 sin 30⁰ , 2 cos 30⁰) = (–1,$\sqrt{3}$)

Hence, equation of BD is

y – 1 = $\frac{\sqrt{3} - 1}{- (\sqrt{3} + 1)}$ (x – $\sqrt{3}$)

= ($\sqrt{3}$– 2) (x – $\sqrt{3}$)

$\left\lbrack \frac{(\sqrt{3} - 1)^{2}}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{4 - 2\sqrt{3}}{2} = 2–\sqrt{3} \right\rbrack$

i.e. (2 – $\sqrt{3}$) x + y = 2$\sqrt{3}$ – 2.