Let \stackrel{\^}{a}= \stackrel{\^}{{i}-2 \stackrel{\^}{j}}/{\sqrt{5}} and \stackrel{\^}{b}= \stackrel{\^}{2 {i}+\stackrel{\^}{j}+3 \stackrel{\^}{k}}/{\sqrt{14}} be two vectors, then the value of (2 \stackrel{\^}{a}+\stackrel{\^}{b}) \cdot[(\stackrel{\^}{a} \times \stackrel{\^}{b}) \times(\stackrel{\^}{a}-2 \stackrel{\^}{b})] is:

NIMCET 2023 Question Paper with solutions

Question : 29 of 120

Marks: +1, -0

Let

{a}↖{\^}=\;{{i}↖{\^}-2 {j}↖{\^}}/{{√{5}}}

and

{b}↖{\^}=\;{2 {i}↖{\^}+{j}↖{\^}+3 {k}↖{\^}}/{{√{14}}}

be two vectors, then the value of

(2 {a}↖{\^}+{b}↖{\^}) ⋅[({a}↖{\^} × {b}↖{\^}) ×({a}↖{\^}-2 {b}↖{\^})]

is:

Solution:

-(2 a+b) ⋅[(a-2 b) ×(a × b)]

\;=-(2 a+b)-[\{(a-2 b) ⋅ b\} {a}↖{→}-\{(a-2 b) ⋅ {a}↖{→}\} {b}↖{→}]

\;=(2 a+b)[\{(a-2 b) ⋅ a\} {b}↖{→}-\{(a-2 b) ⋅ b\} {a}↖{→}]

\;=(2 a+b) [ \{|a|^2-2 b ⋅ a\} {b}↖{→}- \{a ⋅ b-2 b|^2\} {a}↖{→}]

Now,

\;{a}↖{→} ⋅ {b}↖{→}= (\;{{i}↖{\^}-2 {j}↖{\^}}/{{√{5}}}) ⋅ (\;{2 {i}↖{\^}+{j}↖{\^}+3 {k}↖{\^}}/{{√{14}}}) =2-2=0

\;|{a}↖{→}|=1,|{b}↖{→}|=1

\;=(2 a+b) ⋅[\{1-0\} {b}↖{→}-\{0-2\} {a}↖{→}]

\;=(2 a+b) ⋅[{b}↖{→}+2 {a}↖{→}]

\;=(2 a+b) ⋅[b+2 a]

\;=2 a ⋅ b+4|a|^2+|b|^2+4 {a}↖{→} ⋅ {b}↖{→}

\;=0+4+1+0=5 .

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