NIMCET 2025 — Mathematics PYQ

1. Square the given magnitude: The magnitude of the sum of the vectors is given as $|\vec{a}+\vec{b}|=1$. Squaring both sides yields $|\vec{a}+\vec{b}{|}^2=1^2=1$.

2. Expand the squared magnitude: The squared magnitude can be expanded using the dot product property:
   $|\vec{a}+\vec{b}{|}^2=(\vec{a}+\vec{b})\cdot (\vec{a}+\vec{b})=\vec{a}\cdot \vec{a}+2(\vec{a}\cdot \vec{b})+\vec{b}\cdot \vec{b}$.
   This simplifies to $|\vec{a}{|}^2+2(\vec{a}\cdot \vec{b})+|\vec{b}{|}^2=1$.

3. Substitute known values: Given $|\vec{a}|=3$ and $|\vec{b}|=4$, these values are substituted into the equation:
   $3^2+2(\vec{a}\cdot \vec{b})+4^2=1$.
   This simplifies to $9+2(\vec{a}\cdot \vec{b})+16=1$.

4. Solve for the dot product: The equation becomes $25+2(\vec{a}\cdot \vec{b})=1$. Subtracting $25$ from both sides gives
   $2(\vec{a}\cdot \vec{b})=1-25=-24$.
   Dividing by $2$ yields $\vec{a}\cdot \vec{b}=-12$.

5. Consider the magnitude of the difference: The magnitude of the difference of the vectors is expressed as $|\vec{a}-\vec{b}|$. Squaring this magnitude gives
   $|\vec{a}-\vec{b}{|}^2=(\vec{a}-\vec{b})\cdot (\vec{a}-\vec{b})=\vec{a}\cdot \vec{a}-2(\vec{a}\cdot \vec{b})+\vec{b}\cdot \vec{b}$.
   This simplifies to $|\vec{a}{|}^2-2(\vec{a}\cdot \vec{b})+|\vec{b}{|}^2$.

6. Substitute values into the difference equation: The known values $|\vec{a}|=3$, $|\vec{b}|=4$, and $\vec{a}\cdot \vec{b}=-12$ are substituted:
   $|\vec{a}-\vec{b}{|}^2=3^2-2(-12)+4^2$.

7. Calculate the squared magnitude of the difference: This calculation results in
   $|\vec{a}-\vec{b}{|}^2=9+24+16=49$.

8. Find the magnitude of the difference: Taking the square root of both sides gives
   $|\vec{a}-\vec{b}|=\sqrt{49}=7$. [1]

Final Answer
The value of $|\vec{a}-\vec{b}|$ is $7$.