Question

Find the set of values of $p$ for which the angle between the vectors

$\vec{V_1}=x^2\hat{i}+2p\hat{j}+\frac{1}{2}\hat{k}$

$\vec{V_2}=p\hat{i}+x\hat{j}+\hat{k}$ is acute $\forall x\in R$.

$\Rightarrow \vec{V_1}.\vec{V_2}>0$ - For acute angle

Answer 1

The set of values of $p$ is $[0, \frac{1}{2})$.

Answer 2

The angle between two vectors $\vec{V_1}$ and $\vec{V_2}$ is acute if their dot product is positive, i.e., $\vec{V_1} \cdot \vec{V_2} > 0$.

Given vectors are: $\vec{V_1} = x^2\hat{i} + 2p\hat{j} + \frac{1}{2}\hat{k}$ $\vec{V_2} = p\hat{i} + x\hat{j} + \hat{k}$

Calculate the dot product: $\vec{V_1} \cdot \vec{V_2} = (x^2)(p) + (2p)(x) + \left(\frac{1}{2}\right)(1)$ $\vec{V_1} \cdot \vec{V_2} = px^2 + 2px + \frac{1}{2}$

We need this expression to be strictly positive for all $x \in R$: $px^2 + 2px + \frac{1}{2} > 0 \quad \forall x \in R$

We consider two cases for the coefficient of $x^2$:

Case 1: $p = 0$ If $p=0$, the inequality becomes: $(0)x^2 + 2(0)x + \frac{1}{2} > 0$ $\frac{1}{2} > 0$ This statement is true. Thus, $p=0$ is a valid value for which the angle is acute for all $x \in R$.

Case 2: $p \neq 0$ If $p \neq 0$, the expression $px^2 + 2px + \frac{1}{2}$ is a quadratic in $x$. For a quadratic expression $Ax^2 + Bx + C$ to be strictly positive for all real $x$, two conditions must be met:

1. The leading coefficient $A$ must be positive ($A > 0$).
2. The discriminant $D$ must be negative ($D < 0$), indicating no real roots, so the parabola never touches or crosses the x-axis.

In our quadratic $px^2 + 2px + \frac{1}{2}$, we have $A=p$, $B=2p$, and $C=\frac{1}{2}$.

Condition 1: $p > 0$.

Condition 2: $D < 0$ The discriminant $D$ is $B^2 - 4AC$. $D = (2p)^2 - 4(p)\left(\frac{1}{2}\right)$ $D = 4p^2 - 2p$ We need $D < 0$: $4p^2 - 2p < 0$ Factor out $2p$: $2p(2p - 1) < 0$ To solve this inequality, we find the roots of $2p(2p-1)=0$, which are $p=0$ and $p=\frac{1}{2}$. Since the parabola $4p^2-2p$ opens upwards, $2p(2p-1)$ is negative between its roots. So, $0 < p < \frac{1}{2}$.

Combining Condition 1 ($p>0$) and Condition 2 ($0 < p < \frac{1}{2}$), the intersection is $0 < p < \frac{1}{2}$.

Conclusion: From Case 1, $p=0$ is a valid value. From Case 2, $0 < p < \frac{1}{2}$ are valid values. Combining both cases, the set of values for $p$ is $[0, \frac{1}{2})$.