Question

If \( \bar{a} \) and \( \bar{b} \) are two vectors such that \( |\bar{a}|=5 \), \( |\bar{b}|=12 \) and \( |\bar{a}-\bar{b}|=13 \) then \( |2\bar{a}+\bar{b}| = \)

• A. \( 2\sqrt{61} \)
• B. 15
• C. \( 61\sqrt{2} \)
• D. 17

Hint:

Checking for Pythagorean triplets (like 5, 12, 13) allows you to instantly set the dot product to zero without writing out the expansion formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We determine the angle (or dot product) between the vectors using the given magnitudes and the difference vector, then use it to find the magnitude of the sum vector.
Step 2: Key Formula or Approach:

\( |\vec{u} \pm \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 \pm 2\vec{u}\cdot\vec{v} \).
Step 3: Detailed Explanation:

Given \( |\bar{a}|=5, |\bar{b}|=12, |\bar{a}-\bar{b}|=13 \). Note that \( 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \). This implies vectors \(\bar{a}\) and \(\bar{b}\) form a right-angled triangle with the difference vector as hypotenuse. Thus, they are perpendicular. \( \bar{a} \cdot \bar{b} = 0 \). We need \( |2\bar{a}+\bar{b}| \). Let this be \( X \). \[ X^2 = |2\bar{a}+\bar{b}|^2 = 4|\bar{a}|^2 + |\bar{b}|^2 + 4(\bar{a}\cdot\bar{b}) \] \[ X^2 = 4(25) + 144 + 0 \] \[ X^2 = 100 + 144 = 244 \] \[ X = \sqrt{244} = \sqrt{4 \times 61} = 2\sqrt{61} \]
Step 4: Final Answer:

The magnitude is \( 2\sqrt{61} \).