Question

Let $\overrightarrow{x}$, $\overrightarrow{y}$ are non zero and non collinear vectors such that $(a\alpha^2 + b\alpha + c)\overrightarrow{x} + (a\beta^2 + b\beta + c)\overrightarrow{y} + (a\gamma^2 + b\gamma + c)(\overrightarrow{x}\times\overrightarrow{y}) = 0$, then $(a + b + c)$ is (where a,b,c $\in$ R)

• A. -2
• B. -1
• C. 0
• D. 4

Answer 1

Answer: (C)

0

Answer 2

The given vector equation is a linear combination of vectors $\overrightarrow{x}$, $\overrightarrow{y}$, and $\overrightarrow{x}\times\overrightarrow{y}$ equal to the zero vector. Since $\overrightarrow{x}$ and $\overrightarrow{y}$ are non-zero and non-collinear, the set of vectors $\{\overrightarrow{x}, \overrightarrow{y}, \overrightarrow{x}\times\overrightarrow{y}\}$ is linearly independent. Thus, the coefficients in the linear combination must all be zero.

Let $P(t) = at^2 + bt + c$. The coefficients are $P(\alpha)$, $P(\beta)$, and $P(\gamma)$. So, we have $P(\alpha) = 0$, $P(\beta) = 0$, and $P(\gamma) = 0$. This means that $\alpha$, $\beta$, and $\gamma$ are roots of the polynomial equation $at^2 + bt + c = 0$.

A polynomial of degree at most 2 ($at^2+bt+c$) can have at most two distinct roots unless it is the zero polynomial ($a=0, b=0, c=0$). The problem asks for a unique value of $a+b+c$.

If $\alpha, \beta, \gamma$ are distinct, then the only way for them to be roots is if $a=0, b=0, c=0$. In this case, $a+b+c = 0$.

If $\alpha, \beta, \gamma$ are not distinct, the equations $P(\alpha)=0, P(\beta)=0, P(\gamma)=0$ might have non-zero solutions for $a, b, c$. However, the value of $a+b+c = P(1)$ would not be uniquely determined in general (it would depend on $\alpha, \gamma$ and $a$ if $\alpha=\beta \neq \gamma$, or on $\alpha$ and $a$ if $\alpha=\beta=\gamma$).

The fact that the question asks for a unique value of $a+b+c$ implies that the conditions must force $a=b=c=0$, which occurs when $\alpha, \beta, \gamma$ are distinct.

Assuming $\alpha, \beta, \gamma$ are distinct, we get $a=b=c=0$. Therefore, $a+b+c = 0+0+0 = 0$.