A) (0, 0, 0)
B) (0, 0, 1)
C) (0, - 4, -1)
D) (4, 0, -1)
E) (4, 1, -1)
Correct Answer: D
Solution :
The given lines,\[r=(i+j-k)+\lambda (3i-j)\] ...(i) and \[r=(4i-k)+\mu (2i+3k)\] ...(ii) From Eqs. (i) and (ii), \[(i+j-k)+\lambda (3i-j)\] \[=(4i-k)+\mu (2i+3k)\] \[(1+3\lambda )i+(1-\lambda )j-k\] \[=(4+2\mu )i+0j+(-1+3\mu )k\] Equating the coefficient of\[i,\text{ }j\]and k on both sides, \[1+3\kappa =4+2\mu \] \[3\lambda -2\mu =3\] \[1-\lambda =0\] \[\Rightarrow \] \[\lambda =1\] \[-1+3\mu =-1\] \[\Rightarrow \] \[\mu =0\] On putting these values in Eq. (i), \[r=(i+j-k)+(3i-j)\] \[r=4i+0j-k\] So, the intersection point is\[(4,0,-1)\].
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