Question

A unit vector which make ${45}^0$ with vector $2i+2j-k$ and angle of ${60}^0$ with vector $j-k$ is

• A. $\frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}k$
• B. $\frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}j$
• C. $\frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}j+\frac{1}{\sqrt{2}}j$
• D. none of the above

Answer 1

The correct option is D none of the above

Let $\overrightarrow{A}=a\hat{i}+b\hat{j}+c\hat{k}$ be the required unit vector .

Hence , $A=a^2+b^2+c^2=1$ .................eq1

Now , the angle between $\overrightarrow{A}$ and $2\hat{i}+2\hat{j}-\hat{k}$ (magnitude $=3$) is ${45}^o$ ,

therefore , by $\overrightarrow{A}.\overrightarrow{B}=AB\cos \theta$

$(2\hat{i}+2\hat{j}-\hat{k}).\overrightarrow{A}=1\times 3\times \cos 45$

or $(2\hat{i}+2\hat{j}-\hat{k}).(a\hat{i}+b\hat{j}+c\hat{k})=1\times 3\times \cos 45$

or $2a+2b-c=3/\sqrt{2}$ ...............................eq2

And ,the angle between $\overrightarrow{A}$ and $\hat{j}-\hat{k}$ (magnitude $=\sqrt{2}$) is ${60}^o$ ,

therefore , by $\overrightarrow{A}.\overrightarrow{B}=AB\cos \theta$

$(\hat{j}-\hat{k}).\overrightarrow{A}=1\times \sqrt{2}\times \cos 60$

or $(\hat{j}-\hat{k}).(a\hat{i}+b\hat{j}+c\hat{k})=1\times \sqrt{2}\times \cos 60$

or $b-c=1/\sqrt{2}$

or $c=b-1/\sqrt{2}$ ...............................eq3

subtracting eq3 from eq2 , we get

$2a+b=\sqrt{2}$

or $a=(\sqrt{2}-b)/2$.....................................eq4

putting the value of $c$ and $a$ from eq3 and eq4 , into eq1 , we get

$\left[\right(\sqrt{2}-b)/2{]}^2+b^2+[b-1/\sqrt{2}{]}^2=1$

or $b=2\sqrt{2}/3$ .............................eq5

putting the value of $b$ from eq5 , into eq3 and eq4 , we get

$a=1/3\sqrt{2}$

and $c=1/3\sqrt{2}$

Hence , required vector $\overrightarrow{A}=\hat{i}/3\sqrt{2}+2\sqrt{2}b/3\hat{j}+\hat{k}/3\sqrt{2}$