Question

What is the interior acute angle of the parallelogram whose sides are represented by the vectors $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\hat{k}$ and $\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}+\hat{k}$?

• A. ${60}^o$
• B. ${45}^o$
• C. ${30}^o$
• D. ${15}^o$

Answer 1

Answer: (A)

${60}^o$

Given vectors

$\overrightarrow{a}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\hat{k}$

$\overrightarrow{b}=\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}+\hat{k}$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\hat{k})\cdot (\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}+\hat{k})$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta =\frac{1}{2}-\frac{1}{2}+1$

$\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2+1^2}\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+{(-\frac{1}{\sqrt{2}})}^2+1^2}\cos \theta =1$

$\sqrt{\frac{1}{2}+\frac{1}{2}+1}\sqrt{\frac{1}{2}+\frac{1}{2}+1}\cos \theta =1$

$\sqrt{2}\sqrt{2}\cos \theta =1$

$\Rightarrow 2\cos \theta =1$

$\Rightarrow \cos \theta =\frac{1}{2}$

$\Rightarrow \cos \theta =\cos 60$

$\Rightarrow \theta =60$

Hence option A is correct