A unit vector in the XOY- plane that makes an angle 30° with the vector {i}+{j} makes an angle 60° with {i}-{j} is

Correct Answer is : 14[(6+2)i^+(6−2)j^]1/4[({6}+{2}){i}+({6}-{2}){j}]41[(6+2)i^+(6−2)j^]

Correct Answer is : 14[(6+2)i^+(6−2)j^]1/4[({6}+{2}){i}+({6}-{2}){j}]41[(6+2)i^+(6−2)j^]

Test Index

KEAM 2011 Math Paper

Question : 25 of 120

Marks: +1, -0

\hat{i}+\hat{j}

\hat{i}-\hat{j}

$\frac{1}{4}[(\sqrt{6}+\sqrt{2})\hat{i}-(\sqrt{6}-\sqrt{2})\hat{j}]$
$\frac{1}{2}[(\sqrt{6}-\sqrt{2})\hat{i}+(\sqrt{6}+\sqrt{2})\hat{j}]$
$\frac{1}{4}[(\sqrt{6}-\sqrt{2})\hat{i}+(\sqrt{6}+\sqrt{2})\hat{j}]$
$\frac{1}{3}[(\sqrt{6}+\sqrt{2})\hat{i}+(\sqrt{2}-\sqrt{6})\hat{j}]$
$\frac{1}{4}[(\sqrt{6}+\sqrt{2})\hat{i}+(\sqrt{6}-\sqrt{2})\hat{j}]$

Solution:

Let a unit vector in

X O Y

-plane is,

\hat{a}=\frac{a}{|a|}

a=a_{1}i+a_{2}j,|a|=\sqrt{a_{1}^{2}+a_{2}^{2}}

Given vectors let

b=i+j

and

c=i-j

From question;

\hat{a}\cdot b=|\hat{a}|\cdot|b|\cos30^{\circ}

=1\cdot\sqrt{2}\cdot\frac{\sqrt{3}}{2}\Rightarrow\frac{\sqrt{6}}{2}

\Rightarrow\frac{a\cdot b}{|a|}=\frac{\sqrt{6}}{2}

\Rightarrow a_{1}+a_{2}=\frac{\sqrt{6}}{2}|a|

.....(i)

and

\hat{a}\cdot c=|\hat{a}||c|\cdot\cos60^{\circ}

=1\cdot\sqrt{2}\cdot\frac{1}{2}

\frac{a\cdot c}{|a|}=\frac{\sqrt{2}}{2}

\Rightarrow a_{1}-a_{2}=\frac{\sqrt{2}}{2}|a|

.....(ii)

From Eqs. (i) and (ii),

2a_{1}=(\sqrt{6}+\sqrt{2})\frac{|a|}{2}

a_{1}=\frac{1}{4}(\sqrt{6}+\sqrt{2})|a|

\Rightarrow\frac{a_{1}}{|a|}=\frac{1}{4}(\sqrt{6}+\sqrt{2})

and

a_{2}=\frac{1}{4}(\sqrt{6}-\sqrt{2})|a|

\Rightarrow\frac{a_{2}}{|a|}=\frac{1}{4}(\sqrt{6}-\sqrt{2})

So, the unit vector is

\hat{a}=\frac{1}{|a|}(a_{1}i+a_{2}j)

\hat{a}=\frac{1}{4}\{(\sqrt{6}+\sqrt{2})i+(\sqrt{6}-\sqrt{2})j\}

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