The sum of two vectors \overrightarrow{a} and \overrightarrow{b} is a vector \overrightarrow{c} such that |\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=2. Then, the magnitude of \overrightarrow{a}-\overrightarrow{b} is equal to:

NIMCET 2014 Question Paper with solutions

Question : 21 of 120

Marks: +1, -0

The sum of two vectors

{a}↖{→}

and

{b}↖{→}

is a vector

{c}↖{→}

such that

|{a}↖{→}|=|{b}↖{→}|=|{c}↖{→}|=2

. Then, the magnitude of

{a}↖{→}-{b}↖{→}

is equal to:

Solution:

We are given that "sum of two vectors

{a}↖{→}

and

{b}↖{→}

is a vector

{c}↖{→}

⇒ {a}↖{→}+{b}↖{→}={c}↖{→}

Taking dot product of both sides with themselves, the magnitudes will still be equal:

⇒({a}↖{→}+{b}↖{→}) ⋅({a}↖{→}+{b}↖{→})=({c}↖{→}) ⋅({c}↖{→})

⇒|{a}↖{→}|^{2}+|{b}↖{→}|^{2}+2 {a}↖{→} ⋅ {b}↖{→}=|{c}↖{→}|^{2}

Since

|{ a }↖{→}|=|{ b }↖{→}|=|{ c }↖{→}|=2

, we get:

⇒ 2^{2}+2^{2}+2 { a }↖{→} ⋅ { b }↖{→}=2^{2}

⇒ 4+4+2 {a}↖{→} ⋅ {b}↖{→}=4

⇒ 2 {a}↖{→} ⋅ {b}↖{→}=-4

Now,

|{a}↖{→}-{b}↖{→}|^{2}=({a}↖{→}-{b}↖{→}) ⋅({a}↖{→}-{b}↖{→})

=|{a}↖{→}|^{2}+|{b}↖{→}|^{2}-2 {a}↖{→} ⋅ {b}↖{→}

=4+4-(-4)

=12

⇒|{a}↖{→}-{b}↖{→}|=√{12}=2 √{3} .

Go to Question:

Prev Question

Next Question