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NCERT Exemplar · Q7

Q.Find the equations of the two lines through the origin which intersect the line x−32=y−31=z1\dfrac{x-3}{2} = \dfrac{y-3}{1} = \dfrac{z}{1} at angles of π3\dfrac{\pi}{3} each.

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A line through the origin meeting the given line at P=(3+2t,3+t,t)P=(3+2t,3+t,t) at 60∘60^\circ forces t2+3t+2=0t^2+3t+2=0, so t=−1,−2t=-1,-2, giving directions (1,2,−1)(1,2,-1) and (1,−1,2)(1,-1,2): the lines x1=y2=z−1\tfrac{x}{1}=\tfrac{y}{2}=\tfrac{z}{-1} and x1=y−1=z2\tfrac{x}{1}=\tfrac{y}{-1}=\tfrac{z}{2}.

The idea (do not forget the intersection condition)

The required line must (i) pass through the origin, (ii) actually intersect the given line, at (iii) an angle of π3\tfrac{\pi}{3}. The intersection condition is essential: without it the angle alone gives a whole cone of directions. The clean way to build in intersection is to let the line pass through a general point PP of the given line, so the required line is simply OPOP.

Set up

Write the given line in parameter form. With

x−32=y−31=z1=t,\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=t,

a general point is

P=(3+2t, 3+t, t),P = (3+2t,\ 3+t,\ t),

and the given line's direction is d⃗=(2,1,1)\vec{d} = (2,1,1).

The line through the origin and PP has direction OP⃗=(3+2t, 3+t, t)\vec{OP} = (3+2t,\,3+t,\,t).

Apply the angle condition

We need the angle between OP⃗\vec{OP} and d⃗\vec{d} to be π3\tfrac{\pi}{3}:

cos⁡π3=12=∣OP⃗⋅d⃗∣∣OP⃗∣ ∣d⃗∣.\cos\frac{\pi}{3} = \frac{1}{2} = \frac{|\vec{OP}\cdot\vec{d}|}{|\vec{OP}|\,|\vec{d}|}.

Compute each piece:

OP⃗⋅d⃗=2(3+2t)+(3+t)+t=9+6t,\vec{OP}\cdot\vec{d} = 2(3+2t) + (3+t) + t = 9 + 6t,

∣d⃗∣=6,|\vec{d}| = \sqrt{6},

∣OP⃗∣2=(3+2t)2+(3+t)2+t2=6t2+18t+18.|\vec{OP}|^2 = (3+2t)^2 + (3+t)^2 + t^2 = 6t^2 + 18t + 18.

So

∣9+6t∣6 6t2+18t+18=12.\frac{|9+6t|}{\sqrt{6}\,\sqrt{6t^2+18t+18}} = \frac{1}{2}.

Solve for tt

Square both sides:

4(9+6t)2=6 (6t2+18t+18).4(9+6t)^2 = 6\,(6t^2+18t+18). …

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