Question

The plane ax + by = 0 is rotated through an angle a about its line of intersection with the plane z = 0. The equation of the plane in new position is –

• A. ax + by ± z$\sqrt{a^{2} + b^{2}}$ tan a = 0
• B. (ax + by)$\sqrt{a^{2} + b^{2}}$± z tan a = 0
• C. tana (ax + by) ± z$\sqrt{a^{2} + b^{2}}$ = 0
• D. ax + by ±$\sqrt{a^{2} + b^{2}}$z = tan a

Answer 1

Answer: (A)

ax + by ± z$\sqrt{a^{2} + b^{2}}$ tan a = 0

Answer 2

Given planes are ax + by = 0 … (i)

and z = 0 … (ii)

\ Equation of any plane passing through the line of

intersection of planes (i) and (ii) may be taken as ;

ax + by + lz = 0 … (iii)

The direction cosines of a normal to the plane (iii) are :

$\frac{a}{\sqrt{a^{2} + b^{2} + \lambda^{2}}}$,$\frac{b}{\sqrt{a^{2} + b^{2} + \lambda^{2}}}$,$\frac{\lambda}{\sqrt{a^{2} + b^{2} + \lambda^{2}}}$

The direction cosines of a normal to the plane (i) are :

$\frac{a}{\sqrt{a^{2} + b^{2}}}$ . $\frac{b}{\sqrt{a^{2} + b^{2}}}$ = 0

Since the angle between the plane (i) and (iii) is a,

cos a = $\frac{a.a + b.b + \lambda.0}{\sqrt{a^{2} + b^{2} + \lambda^{2}}\sqrt{a^{2} + b^{2}}}$

= $\sqrt{\frac{a^{2} + b^{2}}{a^{2} + b^{2} + \lambda^{2}}}$

Ž l² cos² a = a² (1 – cos² a) + b² (1 – cos² a)

Ž l² = $\frac{(a^{2} + b^{2})\sin^{2}\alpha}{\cos^{2}\alpha}$

Ž l = ± $\sqrt{a^{2} + b^{2}}$tan a = 0.