Question

For any three vectors \[\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,,\overset{\to }{\mathop{c}}\,\], prove that \[{{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\left( \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{a}}\, \right)\]

Answer 1

Hint: In this problem, we have to prove the given vectors, which is in whole square format. We can solve this by using the algebraic formula for the whole square of three terms. We know the algebraic formula for the whole square of three terms, we can expand using the algebraic formula to get the proof verified. We should also keep in mind that the vector symbol is very important.

Complete step by step solution:
We know that the given whole square of three vectors which we have to prove is,
\[{{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\left( \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{a}}\, \right)\]
We know that the algebraic formula for whole square of three term which we can use in this problem to prove the given equation of vector is,
\[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab\times bc\times ca \right)\]
We can apply the above whole square of three terms algebraic formula for the left-hand side of the given equation to get the right-hand side.
We can take the left-hand side and apply the formula, we get
\[\Rightarrow LHS={{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}\]
\[\Rightarrow LHS={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\left( \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{a}}\, \right)\]
\[\Rightarrow LHS=RHS\]
Hence the given vector equation \[{{\left| \overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\, \right|}^{2}}={{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{c}}\, \right|}^{2}}+2\left( \overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{a}}\, \right)\] is proved.

Note: Students make mistakes in writing the correct formula for the given problem. We should know some basic algebraic formulas or identities to solve these types of problems. We should also keep in mind that the vector symbol is very important. We know the algebraic formula for the whole square of three terms, we can expand using the algebraic formula to get the proof verified.