Question

Find the equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at an angle of $\frac{\pi }{3}$

• A. $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ or $\frac{x}{1}=\frac{y}{1}=\frac{z}{2}$
• B. $\frac{x}{-1}=\frac{y}{2}=\frac{z}{-1}$ or $\frac{x}{1}=\frac{y}{1}=\frac{z}{-2}$
• C. $\frac{x}{1}=\frac{y}{-2}=\frac{z}{-1}$ or $\frac{x}{-1}=\frac{y}{1}=\frac{z}{2}$
• D. $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ or $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$

Answer 1

The correct option is D $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ or $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$

Let point on line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\lambda$ ... (1)
are $(3+2\lambda ,3+\lambda ,\lambda )$
Equation of line which pass through origin is
$\frac{x-0}{3+2\lambda }=\frac{y-0}{3+\lambda }=\frac{z-0}{\lambda }$ ... (2)
Angle between (1) & (2) is
$\cos \frac{\pi }{3}=\frac{(3+2\lambda )2+(3+\lambda )1+\lambda \times 1}{\sqrt{(3+2\lambda {)}^2+(3+\lambda {)}^2+{\lambda }^2}\sqrt{2^2+1^2+1}}$
On solving, we get
${\lambda }^2+3\lambda +2=0$
$\Rightarrow \lambda =-1,-2$
Putting the value of $\lambda$ in equation (2), we get
$\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ or $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$