a→,b→\overset{\rightarrow}{a}, \overset{\rightarrow}{b}a→,b→ എന്നിവ രണ്ടു സദിശങ്ങളാണ്∣a→∣=2,∣b→∣=3,a→.b→=4|\overset{\rightarrow}{a}|=2, |\overset{\rightarrow}{b}|=3, \overset{\rightarrow}{a}.\overset{\rightarrow}{b}=4∣a→∣=2,∣b→∣=3,a→.b→=4 ആയാൽ ∣a→−b→∣|\overset{\rightarrow}{a}-\overset{\rightarrow}{b}|∣a→−b→∣കണ്ടുപിടിക്കുക . A√6B√3C√5D2Answer: C. √5 Read Explanation: ∣a→−b→∣2=(a→−b→).(a→−b→){|\overset{\rightarrow}{a}-\overset{\rightarrow}{b}|}^2=(\overset{\rightarrow}{a}-\overset{\rightarrow}{b}).(\overset{\rightarrow}{a}-\overset{\rightarrow}{b})∣a→−b→∣2=(a→−b→).(a→−b→)=a→.a→−a→.b→−b→.a→+b→.b→=\overset{\rightarrow}{a}.\overset{\rightarrow}{a}-\overset{\rightarrow}{a}.\overset{\rightarrow}{b}-\overset{\rightarrow}{b}.\overset{\rightarrow}{a}+\overset{\rightarrow}{b}.\overset{\rightarrow}{b}=a→.a→−a→.b→−b→.a→+b→.b→=∣a2∣→−2(a→.b→)+∣b2∣→=\overset{\rightarrow}{|a^2|}-2(\overset{\rightarrow}{a}.\overset{\rightarrow}{b})+\overset{\rightarrow}{|b^2|}=∣a2∣→−2(a→.b→)+∣b2∣→=22−2×4+32=2^2-2\times4+3^2=22−2×4+32=4−8+9=4-8+9=4−8+9=5=5=5∣a→−b→∣=5|\overset{\rightarrow}{a}-\overset{\rightarrow}{b}|=\sqrt5∣a→−b→∣=5 Read more in App