Question

The unit vector in $ZOX-$ plane and making angle $45°$ and $60°$, respectively with $\stackrel{\rightharpoonup }{a}=2\hat{i}+2\hat{j}-\hat{k}$ and $\stackrel{\rightharpoonup }{b}=\hat{j}-\hat{k}$is?

• A. $\frac{-1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$
• B. $\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{k}$
• C. $\frac{1}{3\sqrt{2}}\hat{i}+\frac{4}{3\sqrt{2}}\hat{j}+\frac{1}{3\sqrt{2}}\hat{k}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{k}$

Explanation for the correct option:

Given that $\stackrel{\rightharpoonup }{a}=2\hat{i}+2\hat{j}-\hat{k}$ and $\stackrel{\rightharpoonup }{b}=\hat{j}-\hat{k}$

Let $\stackrel{\rightharpoonup }{r}$ be the required unit vector

As $\stackrel{\rightharpoonup }{r}$ lies in the $ZOX-$plane it is of the form $\stackrel{\rightharpoonup }{r}=x\hat{i}+z\hat{k}$ and $\left|\stackrel{\rightharpoonup }{r}\right|=1$ since it is a unit vector.

The dot product of two vectors $\stackrel{\rightharpoonup }{p}$ and $\stackrel{\rightharpoonup }{q}$ is given as

$\stackrel{\rightharpoonup }{p}.$$\stackrel{\rightharpoonup }{q}$$=\left|\stackrel{\rightharpoonup }{p}\right|\left|\stackrel{\rightharpoonup }{q}\right|\cos \left(\theta \right)$ where $\theta$ is the angle between the two vectors

Hence, the dot product of $\stackrel{\rightharpoonup }{r}$ and $\stackrel{\rightharpoonup }{a}$ is given as

$\stackrel{\rightharpoonup }{r}.$$\stackrel{\rightharpoonup }{a}$$=\left|\stackrel{\rightharpoonup }{r}\right|\left|\stackrel{\rightharpoonup }{a}\right|\cos \left(45°\right)$

$\Rightarrow$$\left(x\hat{i}+z\hat{k}\right).\left(2\hat{i}+2\hat{j}-\hat{k}\right)=1\times \sqrt{2^2+2^2+{(-1)}^2}\times \cos \left(45°\right)$

$\Rightarrow$ $x\times 2+0-z\times 1=1\times \sqrt{9}\times \frac{1}{\sqrt{2}}$

$\Rightarrow$ $2x-z=\frac{3}{\sqrt{2}}$ $...\left(i\right)$

Similarly the dot product of $\stackrel{\rightharpoonup }{r}$ and $\stackrel{\rightharpoonup }{b}$ is given as

$\Rightarrow$ $\stackrel{\rightharpoonup }{r}.$$\stackrel{\rightharpoonup }{b}$$=\left|\stackrel{\rightharpoonup }{r}\right|\left|\stackrel{\rightharpoonup }{b}\right|\cos \left(60°\right)$

$\Rightarrow$$\left(x\hat{i}+z\hat{k}\right).\left(\hat{j}-\hat{k}\right)=1\times \sqrt{1^2+{(-1)}^2}\times \cos \left(60°\right)$

$\Rightarrow$ $z\times \left(-1\right)=1\times \sqrt{2}\times \frac{1}{2}$

$\Rightarrow$ $z=\frac{-1}{\sqrt{2}}$

Substituting this value of $z=\frac{-1}{\sqrt{2}}$ in $\left(i\right)$ we get

$\Rightarrow$$2x-\left(\frac{-1}{\sqrt{2}}\right)=\frac{3}{\sqrt{2}}$

$\Rightarrow$ $2x=\frac{3}{\sqrt{2}}-\frac{1}{\sqrt{2}}$

$\Rightarrow$ $2x=\frac{2}{\sqrt{2}}$

$\Rightarrow$ $x=\frac{1}{\sqrt{2}}$

Thus the required unit vector is $\stackrel{\rightharpoonup }{r}=\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{k}$.

Hence, option (B) i.e. $\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{k}$ is the correct option