If the lines 1x−1=2y−2=1z+3 and 2x−a=3y+2=1z−3 intersect at point P, then the distance of P from the plane z=a is:
Explanation
Solution: L1 par point: (r+1,2r+2,r−3) L2 par point: (2k+a,3k−2,k+3) z-coordinate barabar karne par: r−3=k+3⇒r=k+6 y-coordinate barabar karne par: 2(k+6)+2=3k−2⇒2k+14=3k−2⇒k=16,r=22. Point P ke coordinate: (23,46,19). x-coordinate se a nikalne par: 23=2(16)+a⇒23=32+a⇒a=−9. Distance of P(23,46,19) from z=−9 is ∣19−(−9)∣=28.
Sahi Option: (1)
Explanation
Solution: L1 par point: (r+1,2r+2,r−3) L2 par point: (2k+a,3k−2,k+3) z-coordinate barabar karne par: r−3=k+3⇒r=k+6 y-coordinate barabar karne par: 2(k+6)+2=3k−2⇒2k+14=3k−2⇒k=16,r=22. Point P ke coordinate: (23,46,19). x-coordinate se a nikalne par: 23=2(16)+a⇒23=32+a⇒a=−9. Distance of P(23,46,19) from z=−9 is ∣19−(−9)∣=28.
Sahi Option: (1)
