Question

The magnitude of dot and cross product of two vectors are 6\3and6 respectively. The angle between them will be: a) 0 b) 30 c) 45 d) 50

Answer 1

Correct Question:

The magnitude of dot and cross product of two vectors are 6√(3) and 6 respectively. The angle between them will be: a) 0 b) 30 c) 45 d) 50

Answer:

• The angle (θ) between them is 30°

Given:

1. Dot product = 6√(3)
2. Cross product = 6

Explanation:

$\rule{300}{1.5}$

Now, we know from the Dot product,

$\displaystyle\longrightarrow |\bf{A.B}|=\sf A\; B\;\cos \theta\\\\\\\longrightarrow\sf 6\sqrt{3} = A\; B\;\cos \theta \ \ \ \because\Bigg[|\bf{A.B}| =\sf 6\sqrt{3}\Bigg]\\\\\\\longrightarrow\sf A\; B\;\cos \theta=6\sqrt{3}\\\\\\\longrightarrow\sf A\; B\;\cos \theta=6\sqrt{3} \quad\dfrac{\quad}{}[1]$

Now, from cross product,

$\displaystyle\longrightarrow |\bf{A\times B}|=\sf A\; B\;\sin \theta\\\\\\\longrightarrow\sf 6 = A\; B\;\sin \theta \ \ \ \because\Bigg[|\bf{A \times B}| =\sf 6\Bigg]\\\\\\\longrightarrow\sf A\; B\;\sin \theta=6\\\\\\\longrightarrow\sf A\; B\;\sin \theta=6 \quad\dfrac{\quad}{}[2]$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Dividing equation 2 by 1,

$\displaystyle\longrightarrow\sf \dfrac{A\;B\;\sin\theta}{A\;B\;\cos\theta}=\dfrac{6}{6\sqrt{3}}\\\\\\\longrightarrow\sf \dfrac{\sin\theta}{\cos\theta}=\dfrac{6}{6\sqrt{3}}\\\\\\\longrightarrow\sf \tan\theta=\dfrac{1}{\sqrt{3}}\ \ \ \because\Bigg[ \dfrac{\sin\theta}{\cos\theta}=\tan\theta \Bigg]\\\\\\\longrightarrow\sf \theta =\tan^{-1}\Bigg(\dfrac{1}{\sqrt{3}}\Bigg)\\\\\\\longrightarrow\sf \theta = 30^{\circ}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf \theta = 30^{\circ}}}}}$

∴ The angle (θ) between them is 30°.

$\rule{300}{1.5}$