Question

In 3-D space, let three lines $L_1,L_2$ and $L_3$ be such that
$L_1$ : intersecting the $z$-axis at $P(0,0,2)$ and does not meet the $x$-y plane
$L_2$ : passing through the origin and through the point $P$.
$L_3$ : passing through the origin and making positive angles $(\alpha ,\beta ,\gamma )$ with co-ordinate axes and $45^{\circ }$ angle with line $L_1$
Identify the which of the following statement(s) is(are) correct?

• A. area of the triangle formed by the lines $L_1,L_2$ and $L_3$ is 2 square units.
• B. area of the triangle formed by the lines $L_1,L_2$ and $L_3$ is 8 square units.
• C. If $\beta =60^{\circ }$, then equation of $L_1$ is $x=y;z=2$.
• D. If $\beta =60^{\circ }$ then equation of $L$ is $x+y=0$ and $z=2$

Answer 1

Answer: (A), (C)

$L_1:\frac{x}{a}=\frac{y}{b};z=2$ (parallel to $xy$-plane)
$L_2:z$-axis
$L_3:\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$
$L_3$ is making $45^{\circ }$ with $z$-axis
$\therefore \Delta$ formed by $L_1,L_2,L_3$ is isosceles right angled
$\therefore$ Area of the $\Delta$ is $=\frac{1}{2}\times 2\times 2=2$ square units

Now, $\beta =60^{\circ }\Rightarrow m=\frac{1}{2},n=\frac{1}{\sqrt{2}}\Rightarrow l=\frac{1}{2}$
$\therefore$ dr's of $L_3$ are $1,1,\sqrt{2}$
Applying, angle between $L_1$ and $L_3$
$\cos 45^{\circ }=\frac{a+b+0}{\sqrt{a^2+b^2}\sqrt{1+1+2}}$
$\frac{1}{\sqrt{2}}=\frac{a+b}{2\sqrt{a^2+b^2}}\Rightarrow 2\left(a^2+b^2\right)=(a+b{)}^2\Rightarrow (a-b{)}^2=0\Rightarrow a=b$
$\therefore$ Equation of $L_1$ is : $x=y;z=2$