NIMCET 2023 Question Paper with solutions

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Question : 21 of 120

Marks: +1, -0

Let

{a}↖{→}=2 {i}↖{\^}+{j}↖{\^}-2 {k}↖{\^}

and

{b}↖{→}={i}↖{\^}+{j}↖{\^}

.
Let

{c}↖{→}

be a vector such that

|{c}↖{→}-{a}↖{→}|=3

,

|({a}↖{→} × {b}↖{→}) × {c}↖{→}|=3

and the angle between

{c}↖{→}

and

{a}↖{→} × {b}↖{→}

be

30^{∘}

. Then

{a}↖{→} ⋅ {c}↖{→}

is equal to:

2
$1 \/ 8$
$25 \/ 8$
5

Solution:

{\A} × {\B}={|\table {i}↖{\^} , {j}↖{\^} , {k}↖{\^} ; 2 , 1 , -2 ; 1 , 1 , 0|}

\;\; ={i}↖{\^}(2)-{j}↖{\^}(2)+{k}↖{\^}(1)

\; {\A} × {\B}=2 {i}↖{\^}-2 {j}↖{\^}+{k}↖{\^}

| {\C}- {\A}|=3

| {\C}- {\A}|^2=9

| {\C}|^2+| {\A}|^2-2 {\C} ⋅ {\A}=9

\;⇒| {\C}|^2-2 {\C} ⋅ {\A}=0

...(i)
Now,

|( {\A} × {\B}) × {\C}|=| {\A} × {\B}|| {\C}|\;

\;={√{4+4+1}} ×| {\C}| × sin \;30^{∘}=3

Given,

θ=30^{∘}

⇒ \;\; | {\C}|=2

Put this value in equation (i)
We get,

{ {\C}}↖{→} ⋅ { {\A}}↖{→}=2

.

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