Question

Find the equations of two lines through the origin which intersect the line

Answer 1

A)

Toolbox:

Eqn. of a line is x−a1b1=y−a2b2=z−a3b3=λ where (a1,a2,a3) is a point on the line and (b1,b2,b3) is d.r. of the line.

Angle between two lines whose d.r. are d1 and d2 is cos−1(d1.d2|d1||d2|).

Let the equation of the line be

x−a1b1=y−a2b2=z−a3b3=λ

But given that it passes through origin.

⇒(a1,a2,a3)=(0,0,0)

⇒ the eqn. is xb1=yb2=zb3=λ.......(i)

Let x−32=y−31=z1=μ.........(ii)

point of intersection of (i) and (ii) say Q is given by

Q(2μ+3,μ+3,μ)

d.r.of OQ=(2μ+3,μ+3,μ)

Since (i) passes through origin O,Line (i)= Line OQ

⇒(b1,b2,b3)=(2μ+3,μ+3,μ)

⇒ d.r. of (i)=d1=(b1,b2,b3)=(2μ+3,μ+3,μ)

d.r.of (ii) =d2=(2,1,1)

d1.d2=2(2μ+3)+1(μ+3)+1.μ=6μ+9

|d1|=(2μ+3)2+(μ+3)2+μ2−−−−−−−−−−−−−−−−−−−−√=6μ2+18μ+18−−−−−−−−−−−−√

|d2|=4+1+1−−−−−−−√=6–√

Also given that (i) and (ii) intersect at angle π3

⇒cosπ3 =(d1.d2|d1||d2|)

⇒12=6μ+96μ2+18μ+18−−−−−−−−−−−√6–√

⇒12μ+18=6(μ2+3μ+3−−−−−−−−−√)

→2μ+3=μ2+3μ+3−−−−−−−−−√

squaring both the sides

4μ2+9+12μ=μ2+3μ+3

⇒3μ2+9μ+6=0orμ2+3μ+2=0

⇒μ=−1or−2

⇒d1=(1,2,−1)or(−1,1,−2)

Equation of the required lines are

x1=y2=z−1=λ and

x−1=y1=z−2=λ