NIMCET 2021 Question Paper with solutions

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Question : 27 of 120

Marks: +1, -0

. Angle between

a↖{→}

and

{{b}↖{→}}

is

120° .

If

|{{b}↖{→}}|=2|\α|

and the vector

{{a}↖{→}}+{xb↖{→}}, {{a}↖{→}}-{{b}↖{→}}

are right angles, then

x=

${1}/{3}$
${1}/{5}$
${2}/{3}$
${2}/{5}$

Solution:

Given that,

|{b}|=2|{a}|

.......(1)
and angle between

{a}

and

{b}=120°

Now we have,

{a} \.{b}=|{a}|| {b}| \cos 120°

From (1) we have

a \.b=|a|(2 a|)(-{1}/{2})

a \.b=-|a|^{2}

......(2)
Since

(a+x b)

and

(a-b)

are at right angle

({a}+{x} {b}) \.({a}-{b})=0

|{a}|^{2}-{a} \.{b}+{x} {a} \.{b}-{x}|{b}|^{2}=0

|{a}|^{2}+{a} \.{b}({x}-1)-{x}|{b}|^{2}=0

From (1) & (2), we have

|a|^{2}-(x-1)|a|^{2}-4 x|a|^{2}=0

|a|^{2} \×[1-x+1-4 x]=0

|a|^{2}[2-5 x]=0

|a|^{2} \≠0 . s o, 2-5 x=0

x={2}/{5}

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