∣A1+A2∣=5
∣A1+A2∣2=25
∣A1∣2+∣A2∣2+2A1A2=25
9+25+2A1⋅A2=25
A1.A2=−29
Now, (2A1+3A2)⋅(3A1−2A2)=6∣A1∣2−4A1A2+9A1A2−6∣A2∣2
=6.∣A1∣2+5A1A2−6∣A2∣2
=6×9+5(−29)−6×25
=54−245−150
=−118.5
Let ∣A1∣=3,∣A2∣=5 and ∣A1+A2∣=5. The value of (2A1+3A2)⋅(3A1−2A2) is:
Held on 8 Apr 2019 · Verified 1 Jul 2026.
−112.5
−118.5
−106.5
−99.5
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
A particle of mass 2 kg is projected vertically upward with a speed of 30 m/s. The maximum height reached by the particle is (g = 10 m/s²):
Two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1:x$. The value of $x$ is:
In an experiment the values of two spring constants were measured as $k_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m}$ and $k_{2}=(20 \pm 0.3) \mathrm{N} / \mathrm{m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _____. ($\pi = 22/7$)
Work through every JEE Main Mechanics PYQ, year by year.