Question

The angles of a triangle two of whose sides are represented by the vectors $\sqrt{3}(\overline{a}\times \overline{b})$ and $\overline{b}-(\hat{a}.\overline{b})\hat{a}$ where $\overline{b}$ is a non-zero vector and $\hat{a}$ is a unit vector in the direction of $\overline{a}$ are

• A. ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right);{\tan }^{-1}\left(\frac{1}{2}\right);{\tan }^{-1}\left(\frac{\sqrt{3}+2}{1-2\sqrt{3}}\right)$
• B. ${\tan }^{-1}\left(\sqrt{3}\right);{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right);{\cot }^{-1}\left(0\right)$
• C. ${\tan }^{-1}\left(\sqrt{3}\right);{\tan }^{-1}\left(2\right);{\tan }^{-1}\left(\frac{\sqrt{3}+2}{2\sqrt{3}-1}\right)$
• D. ${\tan }^{-1}\left(1\right);{\tan }^{-1}\left(1\right);{\cot }^{-1}\left(0\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(\sqrt{3}\right);{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right);{\cot }^{-1}\left(0\right)$

Let $ABC$ be a triangle in which the given vectors are represented by the sides $AB$ and $AC.$
$i.e.,AB=\sqrt{3}(a\times b)$
and $AC=b-(a.b)a$
$\therefore AB.AC=\sqrt{3(a\times b)[b-(a.b)a]}$
$=\sqrt{3[(a\times b).b(a.b)(a\times b).a]}=\sqrt{3[0-0]}=0.$
Therefore, $\angle BAC={90}^0$
${AB}^2={\left[\sqrt{3}(a\times b)\right]}^2=3{(a\times b)}^2$
${AC}^2={[b-(a.b)a]}^2$ ...(i)
$={\left(b\right)}^2+{(a.b)}^2{a}^2-2(b.a)(a.b)$
$={\left(b\right)}^2+{(a.b)}^2-2{(a.b)}^2$
$={\left(b\right)}^2-{(a.b)}^2={\left(b\right)}^2={\left|a\right|}^2{\left|b\right|}^2{\cos }^2\theta$
$={\left(b\right)}^2[1-{\left|a\right|}^2{\cos }^2\theta ]={\left(b\right)}^2(1-{\cos }^2\theta )$
$={\left(b\right)}^2{\sin }^2\theta ={\left|a\right|}^2{\left|b\right|}^2{\sin }^2\theta ={(a\times b)}^2$ ...(ii)
Dividing (i) by (ii), we get
$\frac{{AB}^2}{{AC}^2}=\frac{3{(a\times b)}^2}{{(a\times b)}^2}$
$\Rightarrow {AB}^2=3.{AC}^2\Rightarrow AB=\sqrt{3}AC.$
$\tan C=\frac{AB}{AC}=\frac{\sqrt{3}AC}{AC}=\sqrt{3},\therefore \angle C={60}^0$
So $\therefore \angle A=180-{90}^0-{60}^0={30}^0.$
Hence, angles of the triangle are ${30}^0,{90}^0$ and ${60}^0$