If for some \[\alpha \] and \[\beta \] in R, the intersection of the following three planes |
\[x+4y-2z=1\] |
\[x+7y-5z=\beta \] |
\[x+5y+\alpha z=5\] |
is a line in \[{{R}^{3}},\] then \[\alpha +\beta \] is equal to |
A) \[-10\]
B) \[0\]
C) \[2\]
D) 10
Correct Answer: D
Solution :
[d] \[\because \] Three equations have infinitely many solutions, so \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ \alpha =-\,3 \right]\] Putting the value of \[\alpha \] in (3) \[x+4y-2z=1\] and \[x+5y-3z=5\] On solving \[y=z+4\] and \[x=-2z-15\] Substituting these values in (2) \[x+7y-5z=\beta =-2z-15+7z+28-5z\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\left[ \beta =13 \right]\]You need to login to perform this action.
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