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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2012

  • question_answer
        If\[\omega \]is a cube root of unity, then\[\left| \begin{matrix}    1 & \omega  & {{\omega }^{2}}  \\    \omega  & {{\omega }^{2}} & 1  \\    {{\omega }^{2}} & 1 & \omega   \\ \end{matrix} \right|\]is equal to

    A)  1                            

    B)  0    

    C)  \[\omega \]                      

    D)  \[{{\omega }^{2}}\]

    Correct Answer: B

    Solution :

                    Since the plane \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]                                ?..(i) meets\[x-\]axis so put\[y=0\]and\[z=0\]in plane (i), we get\[x=a\] \[\therefore \] This point be\[A(a,0,0)\] Similarly, the plane (i) meets y-axis and z-axis at the point\[B(0,b,0),\]and\[C(0,0,c)\]respectively. Let the equation of the sphere OABC is \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0\]...(ii) Since,   this   equation   passes   through \[O(0,0,0),A(a,0,0),B(0,b,0)\]and\[C(0,0,c),\]therefore \[d=0\]                                   ...(iii) \[{{a}^{2}}+2ua+d=0\]                 ...(iv) \[{{b}^{2}}+2vb+d=0\]                  ...(v) \[{{c}^{2}}+2wc+d=0\]                ...(vi) Putting\[d=0\]from Eq. (iii) in Eqs. (iv), (v) and (vi), we get \[u=-\frac{a}{2},v=-\frac{b}{2},w=-\frac{c}{2}\] Substituting the values of u, v, w and d in Eq. (ii), we get \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-ax-by-cz=0\]which is the required equation of sphere.


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