If ab + bc + ca = 0 and | , * 20 c a - x &c&b & b - x &a &a& c - … - Vidyadip

MCQ

If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, then one of the value of $x$ is

✓
${({a^2} + {b^2} + {c^2})^{\frac{1}{2}}}$
B
${\left[ {\frac{3}{2}({a^2} + {b^2} + {c^2})} \right]^{\frac{1}{2}}}$
C
${\left[ {\frac{1}{2}({a^2} + {b^2} + {c^2})} \right]^{\frac{1}{2}}}$
D
None of these

✓

Answer

Correct option: A.

${({a^2} + {b^2} + {c^2})^{\frac{1}{2}}}$

a
(a) Applying $ {c_1}→{c_1}+{c_2}+{c_3}$

$(a+b+c-x)$ $\left| {\,\begin{array}{*{20}{c}}
1&c&b \\
1&{b - x}&a \\
1&a&{c - x}
\end{array}\,} \right|\, = 0$

=> $(a+b+c-x)$ $[{(b-x)(c-x)-a^2}$ $ + c(a-c+x) + {b({a-b+x)}}] =0$

=> $(a+b+c-x)$ $[(bc-cx+bx+x^2-a^2 + ca - c^2 + cx +ab -b^2 +bx] =0$

==> $(a+b+c)[x^2 -(a^2+b^2+c^2)+ab+bc+ca]=0$

==>$(a+b+c-x)[x^2-(a^2 + b^2 +c^2] = 0$

[ $\because$ $ab + bc + ca = 0$]

$\therefore$ $x=a+b+c $ $ and (a^2+b^2+c^2)^{1/2} $

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