A) \[\left( 0,\,\frac{13}{5},\,2 \right)\]
B) \[\left( 2,\,0,\,\frac{13}{5} \right)\]
C) \[\left( 0,\,2,\,\frac{13}{5} \right)\]
D) (2, 2, 0)
Correct Answer: A
Solution :
Line joining the points (3,5,?7) and (?2,1,8) is, \[\frac{x-3}{(-2)-(3)}=\frac{y-5}{(1)-(5)}=\frac{z-(-7)}{8-(-7)}\] \[\frac{x-3}{-5}=\frac{y-5}{-4}=\frac{z+7}{15}=K\], (Let) ?..(i) \ \[x=-5K+3\], \[y=-4K+5\], \[z=15K-7\] \[\because \] Line (i) meets the yz-plane \[\therefore \] \[-5K+3=0\Rightarrow K=3/5\] Put the value of K in \[x,\,y,\,z\] So the required point is (0, 13/5, 2).
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