Suppose we have two planes $$a_1x+b_1y+c_1z+d_1=0$$ and $$a_2x+b_2y+c_2z+d_2=0$$ where $d_1,d_2 >0 \ or \ <0$ then prove that the angle between planes in which origin lies is acute if $$a_1a_2+b_1b_2+c_1c_2<0$$
The angle between the planes is given by $$cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}} $$
In this the denominator $\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}$ is positive, hence the sign of $cos\theta $ depends only on $a_1a_2+b_1b_2+c_1c_2$
How do I proceed further ?
More precisely, one of the angles between the planes is given by $(1)$.
So, if you mean that $\theta=\text{(the angle between the planes in which origin lies)}$, then we have $$\cos\theta=\color{red}{\pm}\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}} $$
The following will use a different approach.
Let $P_i : a_ix+b_iy+c_iz+d_i=0$, and let $Q_i$ be a point on $P_i$ such that $OQ_i$ is perpendicular to $P_i$. Also, let $\theta$ be the angle between the planes in which origin $O$ lies, and let $\alpha=\angle{Q_1OQ_2}$.
$\qquad\qquad\qquad\qquad\qquad$
Then, since $$OQ_1\ :\ \frac{x}{a_1}=\frac{y}{b_1}=\frac{z}{c_1}$$ we have $$Q_1\left(\frac{-d_1a_1}{a_1^2+b_1^2+c_1^2},\frac{-d_1b_1}{a_1^2+b_1^2+c_1^2},\frac{-d_1c_1}{a_1^2+b_1^2+c_1^2}\right)$$ Similarly, $$Q_2\left(\frac{-d_2a_2}{a_2^2+b_2^2+c_2^2},\frac{-d_2b_2}{a_2^2+b_2^2+c_2^2},\frac{-d_2c_2}{a_2^2+b_2^2+c_2^2}\right)$$
Hence, $$\begin{align}&\text{$\theta$ is acute}\\&\iff \text{$\alpha$ is obtuse}\\&\iff \cos\alpha\lt 0\\&\iff \frac{\vec{OQ_1}\cdot\vec{OQ_2}}{\left|\vec{OQ_1}\right|\cdot\left|\vec{OQ_2}\right|}\lt 0\\&\iff \vec{OQ_1}\cdot\vec{OQ_2}\lt 0\\&\iff \frac{(-d_1a_1)(-d_2a_2)+(-d_1b_1)(-d_2b_2)+(-d_1c_1)(-d_2c_2)}{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}\lt 0\\&\iff d_1d_2(a_1a_2+b_1b_2+c_1c_2)\lt 0\\&\iff a_1a_2+b_1b_2+c_1c_2\lt 0\end{align}$$