The point of intersection of the lines (x-5/3)=(y-7/-1)=(z+2/1) and (x+3/-36)=(y-3/2)=(z-6/4) is

Question

The point of intersection of the lines (x-5/3)=(y-7/-1)=(z+2/1) and (x+3/-36)=(y-3/2)=(z-6/4) is (A) (21, (5/3), (10/3)) (B) (2,10,4) (C) (-3,3,6) (D) (5,7,-2)

Tardigrade Admin · Accepted Answer

Given lines are (x-5/3)=(y-7/-1)=(z+2/1)=r1 (say) and (x+3/-36)=(y-3/2)=(z-6/4)=r2 (say) ∴ x =3 r1+5=-36 r2-3, y =-r1+7=3+2 r2 and z =r1-2=4 r2+6 On solving, we get x=21, y=(5/3), z=(10/3)

Q. The point of intersection of the lines $\frac{x - 5}{3} = \frac{y - 7}{- 1} = \frac{z + 2}{1}$ and $\frac{x + 3}{- 36} = \frac{y - 3}{2} = \frac{z - 6}{4}$ is

$(21, \frac{5}{3}, \frac{10}{3})$

$(2, 10, 4)$

$(- 3, 3, 6)$

$(5, 7, - 2)$

Q. The point of intersection of the lines 3x−5​=−1y−7​=1z+2​ and −36x+3​=2y−3​=4z−6​ is

(21,35​,310​)

(2,10,4)

(−3,3,6)

(5,7,−2)

Given lines are 3x−5​=−1y−7​=1z+2​=r1​ (say) and −36x+3​=2y−3​=4z−6​=r2​ (say) ∴x=3r1​+5=−36r2​−3, y=−r1​+7=3+2r2​ and z=r1​−2=4r2​+6 On solving, we get x=21,y=35​,z=310​

Q. The point of intersection of the lines $\frac{x - 5}{3} = \frac{y - 7}{- 1} = \frac{z + 2}{1}$ and $\frac{x + 3}{- 36} = \frac{y - 3}{2} = \frac{z - 6}{4}$ is

$(21, \frac{5}{3}, \frac{10}{3})$

$(2, 10, 4)$

$(- 3, 3, 6)$

$(5, 7, - 2)$