P,Q,R and S are four points with the position vectors 3i−4j+5k,−4i+5j+kand −3i+4j+3k, respectively. Then, the line PQ meets the line RS at the point
Let the coordinates of four points P,Q,R and S be (3,−4,5),(0,0,4),(−4,5,1) and (−3,4,3)respectively. Now, equation of line PQ is
x−3
0−3
=
y+4
0+4
=
z−5
4−5
⇒
x−3
−3
=
y+4
4
=
z−5
−1
=r1 (say) …(i) Equation of line RS is
x+4
−3+4
=
y−5
4−5
=
z−1
3−1
⇒
x+4
1
=
y−5
−1
=
z−1
2
=r2 (say) …(ii) Let (−3r1+3,4r1−4,−r1+5) and (r2−4,−r2+52r2+1) be the points on line (i) and (ii), respectively. Since, both lines intersect at a common point, then −3r1+3=r2−4 ⇒ 3r1+r2=7 …(iii) and −r2+5=4r1−4 ⇒ 4r1+r2=9 …(iv) On subtracting Eq. (iv) from Eq. (iii), we get r1=2 On putting the value of r1 in Eq. (iii), we get 3(2)+r2=7⇒r2=1 So, required point of intersection is (−3,4,3) i.e., −3i+4j+3k