A) (- 2, 2, 2)
B) \[(1,\,\,1,\,\,\sqrt{10})\]
C) (2, - 2, - 2)
D) None of these
Correct Answer: C
Solution :
| [c] Since, \[\overset{\to }{\mathop{a}}\,\] is \[\bot \] to \[\overset{\to }{\mathop{d}}\,\], so \[x-y+2z=0\] ?(1) |
| Moreover, \[|\overset{\to }{\mathop{b}}\,|=|\overset{\to }{\mathop{c}}\,|,\] so \[\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{c}}\,\] |
| as \[\overset{\to }{\mathop{a}}\,\] makes equal angles with \[\overset{\to }{\mathop{b}}\,\] and \[\overset{\to }{\mathop{c}}\,\]. Thus |
| \[xy-2yz+3xz=2xz+3xy-yz\] |
| \[\Rightarrow xz-2xy-yz=0\] ?(2) |
| Also \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=12\] ?(3) |
| and \[y<0\] |
| Put the value of y from eq. (1) in eq. (2), |
| We get, \[{{x}^{2}}+2xz+{{z}^{2}}=0;\] so, \[x=-z\] and \[y=z\] |
| Again put these values in eq. (3), we get |
| \[{{z}^{2}}=4\Rightarrow z=\pm 2\] |
| But \[y<0\] and \[y=z\]. Hence, \[z=-2=y\] and \[x=2\] |
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