If the vectors $(- b c, b^{2} + b c, c^{2} + b c), (a^{2} + a c, - a c, c^{2} + a c)$ and $(a^{2} + a b, b^{2} + a b, - a b)$ are coplanar, where none of a, b or c is zero, then

a^{2} + b^{2} + c^{2} = 1

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b c + c a + a b = 0

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a + b + c = 0

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a^{2} + b^{2} + c^{2} = b c + c a + a b

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Solution

The correct option is B $b c + c a + a b = 0$
$(- b c {, b}^{2} + b c, c^{2} + b c), (a^{2} + a c, - a c, c^{2} + a c)$ and $(a^{2} + a b, b^{2} + a b, - a b)$
are co-planar when
$∣ ∣ ∣ ∣ ∣ \begin{matrix} - b c & b^{2} + b c & c^{2} + b c a^{2} + a c & - a c & b^{2} + a c a^{2} + a b & b^{2} + a b & - a b \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 \Rightarrow ∣ ∣ ∣ ∣ ∣ \begin{matrix} - b c & b^{2} & c^{2} a^{2} & - a c & c^{2} a^{2} & b^{2} & - a b \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 0 & b c & b c a c & 0 & a c a b & a b & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow b c + a c + a b = 0$

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Similar questions

\frac{a^{2} + b^{2} + a b}{b^{2} + c^{2} + b c} + \frac{c^{2} + c a + a^{2}}{b^{2} + c^{2} + b c}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} + a^{4} & 1 + a b + a^{2} b^{2} & 1 + a c + a^{2} c^{2} 1 + a b + a^{2} b^{2} & 1 + b^{2} + b^{4} & 1 + b c + b^{2} c^{2} 1 + a c + a^{2} c^{2} & 1 + b c + b^{2} c^{2} & 1 + c^{2} + c^{4} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(a - b)}^{2} {(b - c)}^{2} {(c - a)}^{2}

a b + b c + c a = 0

\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - a c} + \frac{1}{c^{2} - a b} = 0

a^{2} + b^{2} + c^{2} - a b - b c - a c = 0

a = 5

b^{2} + c^{2}

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