Question

Find the cosine of the angle between the directions of the vectors
$\overrightarrow{a}=4\hat{i}+3\hat{j}+\hat{k},\overrightarrow{b}=2\hat{i}-\hat{j}+2\hat{k}.$ Also find a unit vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$. What is the sine of the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ ?

• A. $\sin \theta =\sqrt{\left(\frac{185}{26\times 9}\right)}=\frac{1}{3}\sqrt{\left(\frac{185}{26}\right)}.$
• B. $\sin \theta =-\sqrt{\left(\frac{185}{26\times 9}\right)}=-\frac{1}{3}\sqrt{\left(\frac{185}{26}\right)}.$
• C. $\sin \theta =\sqrt{\left(\frac{185}{26\times 8}\right)}$
• D. None of these

Answer 1

The correct option is A $\sin \theta =\sqrt{\left(\frac{185}{26\times 9}\right)}=\frac{1}{3}\sqrt{\left(\frac{185}{26}\right)}.$

$a.b=ab\cos \theta =a_1{a}_2+b_1{b}_2+c_1{c}_2$ where $a=\sqrt{\sum a{{}_1}^2},b=\sqrt{\sum b{{}_1}^2}$

$\therefore \sqrt{16+9+1}\sqrt{4+1+4}\cos \theta =4.2+3(-1)+1.2$

$\Rightarrow \sqrt{26}.3\cos \theta =7\Rightarrow \cos \theta =\frac{7}{3\sqrt{\left(26\right)}}.$

We know that $a\times b$ represents a vector which is perpendicular to both $a$ and $b.$

Now $a\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 3& 1\\ 2& -1& 2\end{array}\right|=7\hat{i}-6\hat{j}-10\hat{k}.$

Hence a unit vector will be obtained by dividing $7\hat{i}-6\hat{j}-10\hat{k}$ by its module $\sqrt{49+36+100}=\sqrt{185}.$

The required unit vector is $\frac{1}{\sqrt{\left(185\right)}}(7\hat{i}-6\hat{j}-10\hat{k})$

Again, $a\times b=ab\sin \theta \hat{n}$

$7i-6j-10k=\sqrt{26}\sqrt{9}\sin \theta \hat{n}.$

Square both sides and we know that square of a vector is equal to square of its module

i.e. $a^2=a^2$ and ${\hat{n}}^2=1$

$\therefore 49+36+100=26\times 9{\sin }^2\theta .1$

$\therefore \sin \theta =\sqrt{\left(\frac{185}{26\times 9}\right)}=\frac{1}{3}\sqrt{\left(\frac{185}{26}\right)}.$