Question

Let \( \vec{a} = 6\hat{i} + 2\hat{j} + 3\hat{k} \). If \( \vec{b} \) is parallel to \( \vec{a} \) and \( \vec{a} \cdot \vec{b} = \frac{49}{2} \), then \( |\vec{b}| = \)

• A. \(49 \)
• B. \(7 \)
• C. \(14 \)
• D. \( 7\sqrt{2} \)
• E. \( \frac{7}{2} \)

Hint:

For parallel vectors, angle = 0 so dot product simplifies to product of magnitudes.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
If two vectors are parallel, one is a scalar multiple of the other. We can use this property along with the given dot product to find the scalar multiple and then the magnitude of the unknown vector.
Step 2: Key Formula or Approach:
1. If \(\vec{b}\) is parallel to \(\vec{a}\), then \(\vec{b} = k\vec{a}\) for some scalar `k`. 2. The dot product \(\vec{a} \cdot \vec{b}\) can be expressed as \(\vec{a} \cdot (k\vec{a}) = k(\vec{a} \cdot \vec{a}) = k|\vec{a}|^2\). 3. The magnitude \(|\vec{b}| = |k\vec{a}| = |k||\vec{a}|\).
Step 3: Detailed Explanation:
1. Find the magnitude of \(\vec{a}\). \[ |\vec{a}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \] So, \(|\vec{a}|^2 = 49\). 2. Use the dot product to find the scalar k. Since \(\vec{b}\) is parallel to \(\vec{a}\), we can write \(\vec{b} = k\vec{a}\). Now, consider the dot product: \[ \vec{a} \cdot \vec{b} = \vec{a} \cdot (k\vec{a}) = k |\vec{a}|^2 \] We are given \(\vec{a} \cdot \vec{b} = \frac{49}{2}\). \[ \frac{49}{2} = k (49) \] Solving for `k`: \[ k = \frac{49/2}{49} = \frac{1}{2} \] 3. Calculate the magnitude of \(\vec{b}\). Now that we know \(k\), we can find the magnitude of \(\vec{b}\): \[ |\vec{b}| = |k\vec{a}| = |k| \cdot |\vec{a}| \] \[ |\vec{b}| = \left|\frac{1}{2}\right| \cdot 7 = \frac{1}{2} \cdot 7 = \frac{7}{2} \] Step 4: Final Answer:
The magnitude of \(\vec{b}\) is \(\frac{7}{2}\).