Question

If the dr's the line are $(1+\lambda ,1-\lambda ,2)$ and it makes an angle ${60}^o$ with the Y-axis then $\lambda$ is

• A. $1\pm \sqrt{3}$
• B. $4\pm \sqrt{5}$
• C. $2\pm 2\sqrt{3}$
• D. $2\pm \sqrt{5}$

Answer 1

The correct option is D $2\pm \sqrt{5}$

Given that,

$\begin{array}{c}\frac{\left|1-\lambda \right|}{\sqrt{{\left(1+\lambda \right)}^2+{\left(1-\lambda \right)}^2+4}}=\frac{1}{2}\\ \frac{\left|1-\lambda \right|}{\sqrt{6+2{\lambda }^2}}=\frac{1}{2}\\ 4{\left(\left|1-\lambda \right|\right)}^2=6+2{\lambda }^2\\ 2{\lambda }^2-2-8\lambda =0\\ {\lambda }^2-4\lambda -1=0\\ \lambda =\frac{4\pm \sqrt{16+4}}{2}=\frac{4\pm 2\sqrt{5}}{2}\\ =2\pm \sqrt{5}\end{array}$

Then,

We get $2\pm \sqrt{5}$

Option $D$ is correct answer.