NIMCET 2023 Question Paper with solutions

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Question : 25 of 120

Marks: +1, -0

If

\cos ^{-1} (\;{3}/{{√{20}}})

is the angle between the vectors

{i}↖{\^}-2 x {j}↖{\^}+3 y {k}↖{\^}

and

x {i}↖{\^}-{j}↖{\^}+y {k}↖{\^}, x, y

are real numbers then which of the following points satisfies the locus of

(x, y)

?

$(0,1)$
$(-1,-1)$
$(1,0)$
$(1,1)$

Solution:

{a}↖{→}={i}↖{\^}-2 x {j}↖{\^}+3 y {k}↖{\^}

and

{b}↖{→}\;=x {i}↖{\^}+{j}↖{\^}+y {k}↖{\^}

\cos θ\;=\;{{a}↖{→} ⋅ {b}↖{→}}/{|{a}↖{→}||{b}↖{→}|}

\;=\cos ^{-1} (\;{3}/{{√{20}}}) =\;{3}/{{√{20}}}

\;{3}/{{√{20}}}\;=\;{x-2 x+2 y^2}/{{√{1+4 x^2+4 y^2}} {√{x^2+1+y^2}}}

⇒ 3 {√{1+4 x^2+4 y^2}} {√{x^2+1+y^2}}\;={√{20}} [2 y^2-x]

From given options

(0,1)

satisfy given equations.

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