Aspire Faculty ID #11962 · Topic: NIMCET 2025 · Just now
NIMCET 2025

If $\vec{a}$ and $\vec{b}$ are twp vectors such that |$\vec{a}$|=3, |$\vec{b}$|=4 and |$\vec{a}+\vec{b}$|=1, then the value of $|\vec{a}-\vec{b}|$ is

Solution

Given: $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{a} + \vec{b}| = 1$. 

Use the identity: $|\vec{a} + \vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} + 2\vec{a}\cdot\vec{b}$ 
Substitute values: $1^{2} = 3^{2} + 4^{2} + 2\vec{a}\cdot\vec{b}$ 
$1 = 9 + 16 + 2\vec{a}\cdot\vec{b}$ 
$1 = 25 + 2\vec{a}\cdot\vec{b}$ 
$2\vec{a}\cdot\vec{b} = -24$ 
$\vec{a}\cdot\vec{b} = -12$ 
Now compute $|\vec{a} - \vec{b}|$: 
$|\vec{a} - \vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} - 2\vec{a}\cdot\vec{b}$ 
$= 9 + 16 - 2(-12)$ 
$= 25 + 24$ $= 49$ 
So: $|\vec{a} - \vec{b}| = 7$

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