Question

If $\vec{a} \cdot \vec{b}=0$ and $\vec{a}+\vec{b}$ makes an angle $60^{\circ}$ with $\vec{a} $ then

• A. $|\vec{a}|=2| \vec{b} \mid$
• B. $2|\vec{a}|=|\vec{b}|$
• C. $|\vec{a}|=\sqrt{3}| \vec{b} \mid$
• D. $\sqrt{3}| \vec{a}|=| \vec{b} \mid$

Answer 1

The Correct Option is D

To solve the problem, we start by interpreting the given conditions and applying relevant vector properties:

1. The condition \(\vec{a} \cdot \vec{b} = 0\)specifies that vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular.
2. Given that the angle between \(\vec{a} + \vec{b}\) and \(\vec{a}\) is \(60^\circ\), we can utilize the cosine rule for vector angles:

For two vectors \(\vec{u}\) and \(\vec{v}\), the cosine of the angle \(\theta\) between them can be given by:

\(\cos \theta = \frac{ \vec{u} \cdot \vec{v} }{|\vec{u}| |\vec{v}|}\)

Setting the vectors as \(\vec{u} = \vec{a} + \vec{b}\) and \(\vec{v} = \vec{a}\), and the angle \(\theta = 60^\circ\), we have:

\(\cos 60^\circ = \frac{ (\vec{a} + \vec{b}) \cdot \vec{a} }{|\vec{a} + \vec{b}| |\vec{a}|}\)

Given \(\cos 60^\circ = \frac{1}{2}\), the equation becomes:

\(\frac{1}{2} = \frac{\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a}}{\sqrt{|\vec{a}|^2 + |\vec{b}|^2} |\vec{a}|}\)

Since \(\vec{b} \cdot \vec{a} = 0\):

\(\frac{1}{2} = \frac{|\vec{a}|^2}{\sqrt{|\vec{a}|^2 + |\vec{b}|^2} |\vec{a}|}\)

Solving for this, we have:

\(\frac{|\vec{a}|}{2} = \frac{1}{\sqrt{|\vec{a}|^2 + |\vec{b}|^2}}\)

\(|\vec{a}|^2 = \frac{1}{2} (|\vec{a}|^2 + |\vec{b}|^2)\)

Simplifying this equation yields:

\(|\vec{b}|^2 = 3|\vec{a}|^2\)

Taking the square root on both sides gives:

\(|\vec{b}| = \sqrt{3} |\vec{a}|\)

1. Thus, the correct answer is the option where \(\sqrt{3}|\vec{a}| = |\vec{b}|\).

Therefore, the correct answer is:

\(\sqrt{3}| \vec{a}|=| \vec{b} \mid\)