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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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