Question

If \(\vec{a}.\vec{b}=0\) and \(\vec{a}+\vec{b}\) makes an angle 60° with \(\vec{a}\) then

• A. \(|\vec{a}|=2|\vec{b}|\)
• B. \(2|\vec{a}|=|\vec{b}|\)
• C. \(|\vec{a}|=\sqrt3|\vec{b}|\)
• D. \(\sqrt3|\vec{a}|=|\vec{b}|\)

Answer 1

The Correct Option is D

Given that \( \mathbf{a} \cdot \mathbf{b} = 0 \), it means \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular to each other. We are also told that \( \mathbf{a} + \mathbf{b} \) makes an angle of 60° with \( \mathbf{a} \). The angle between \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} \) is 60°. Using the formula for the dot product: \[ \cos(60^\circ) = \frac{(\mathbf{a} + \mathbf{b}) \cdot \mathbf{a}}{|\mathbf{a} + \mathbf{b}| |\mathbf{a}|} \] Since \( \cos(60^\circ) = \frac{1}{2} \), we get: \[ \frac{(\mathbf{a} + \mathbf{b}) \cdot \mathbf{a}}{|\mathbf{a} + \mathbf{b}| |\mathbf{a}|} = \frac{1}{2} \] Now calculate \( (\mathbf{a} + \mathbf{b}) \cdot \mathbf{a} \): \[ (\mathbf{a} + \mathbf{b}) \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{a} = |\mathbf{a}|^2 \] because \( \mathbf{a} \cdot \mathbf{b} = 0 \). The magnitude of \( \mathbf{a} + \mathbf{b} \) is: \[ |\mathbf{a} + \mathbf{b}| = \sqrt{|\mathbf{a}|^2 + |\mathbf{b}|^2} \] Substitute these values into the dot product formula: \[ \frac{|\mathbf{a}|^2}{\sqrt{|\mathbf{a}|^2 + |\mathbf{b}|^2} |\mathbf{a}|} = \frac{1}{2} \] Simplifying: \[ \frac{|\mathbf{a}|}{\sqrt{|\mathbf{a}|^2 + |\mathbf{b}|^2}} = \frac{1}{2} \] Squaring both sides: \[ \frac{|\mathbf{a}|^2}{|\mathbf{a}|^2 + |\mathbf{b}|^2} = \frac{1}{4} \] Now, solving for \( |\mathbf{a}|^2 \) and \( |\mathbf{b}|^2 \): \[ 4 |\mathbf{a}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 \] \[ 3 |\mathbf{a}|^2 = |\mathbf{b}|^2 \] Thus: \[ |\mathbf{b}| = \sqrt{3} |\mathbf{a}| \]

The correct answer is (D) : \(\sqrt3|\vec{a}|=|\vec{b}|\).

Answer 2

Given:

• \(\vec{a} \cdot \vec{b} = 0\) (i.e., \(\vec{a}\) and \(\vec{b}\) are perpendicular)
• \(\vec{a} + \vec{b}\) makes an angle of 60° with \(\vec{a}\)

Let:

• \(|\vec{a}| = a\)
• \(|\vec{b}| = b\)

Step 1: Use the cosine formula between two vectors:

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

Step 2: Simplify the numerator using dot product:

\[ (\vec{a} + \vec{b}) \cdot \vec{a} = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} = a^2 + 0 = a^2 \]

Step 3: Find magnitude of \(\vec{a} + \vec{b}\)

Since \(\vec{a} \perp \vec{b}\), we use: \[ |\vec{a} + \vec{b}| = \sqrt{a^2 + b^2} \]

Step 4: Plug into the cosine formula:

\[ \cos(60^\circ) = \frac{a^2}{a\sqrt{a^2 + b^2}} = \frac{1}{2} \]

Step 5: Solve:

\[ \frac{a^2}{a\sqrt{a^2 + b^2}} = \frac{1}{2} \Rightarrow \frac{a}{\sqrt{a^2 + b^2}} = \frac{1}{2} \Rightarrow 2a = \sqrt{a^2 + b^2} \]

Squaring both sides:

\[ 4a^2 = a^2 + b^2 \Rightarrow 3a^2 = b^2 \Rightarrow a = \sqrt{3}b \]

Hence,\[ |\mathbf{b}| = \sqrt{3} |\mathbf{a}| \]

Correct option: \[ |\mathbf{b}| = \sqrt{3} |\mathbf{a}| \]