Question

Given vectors \( \vec{a}, \vec{b}, \vec{c} \) such that \( \vec{a} \) is perpendicular to \( \vec{b} \) and \( \vec{c} \), \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \), and \( \vec{b} \cdot \vec{c} = 1 \). If there is a nonzero vector \( \vec{d} \) coplanar with \( \vec{a} + \vec{b} \) and \( 2\vec{b} - \vec{c} \), and if \( \vec{d} \cdot \vec{a} = 1 \), then calculate \( |\vec{d}|^2 \).

• A. \( 13y^2 + 14y + 5 \)
• B. \( y^2 + 14y + 5 \)
• C. \( y^2 - 14y - 5 \)
• D. \( y^2 - 14y + 5 \)

Hint:

To solve vector problems involving dot products and magnitudes, express the vector as a linear combination of others and apply the given conditions to solve for unknowns.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given properties of vectors \( \bar{\text{a}}, \bar{\text{b}}, \bar{\text{c}} \) and need to find the magnitude squared of vector \( \bar{\text{d}} \) which is defined in terms of a linear combination.
Step 2: Key Formula or Approach:
1. Express \( \bar{\text{d}} = \text{x}(\bar{\text{a}} + \bar{\text{b}}) + \text{y}(2\bar{\text{b}} - \bar{\text{c}}) \).
2. Use the given dot product \( \bar{\text{d}} \cdot \bar{\text{a}} = 1 \) to find \( \text{x} \).
3. Calculate \( |\bar{\text{d}}|^2 = \bar{\text{d}} \cdot \bar{\text{d}} \).
Step 3: Detailed Explanation:
1. \( \bar{\text{d}} = \text{x}\bar{\text{a}} + (\text{x} + 2\text{y})\bar{\text{b}} - \text{y}\bar{\text{c}} \).
2. \( \bar{\text{d}} \cdot \bar{\text{a}} = \text{x}(\bar{\text{a}} \cdot \bar{\text{a}}) + (\text{x} + 2\text{y})(\bar{\text{b}} \cdot \bar{\text{a}}) - \text{y}(\bar{\text{c}} \cdot \bar{\text{a}}) \).
Since \( \bar{\text{a}} \perp \bar{\text{b}} \) and \( \bar{\text{c}} \), then \( \bar{\text{b}} \cdot \bar{\text{a}} = 0 \) and \( \bar{\text{c}} \cdot \bar{\text{a}} = 0 \).
\( 1 = \text{x}(1) \Rightarrow \text{x} = 1 \).
3. So, \( \bar{\text{d}} = \bar{\text{a}} + (1 + 2\text{y})\bar{\text{b}} - \text{y}\bar{\text{c}} \).
4. \( |\bar{\text{d}}|^2 = (\bar{\text{a}} + (1 + 2\text{y})\bar{\text{b}} - \text{y}\bar{\text{c}}) \cdot (\bar{\text{a}} + (1 + 2\text{y})\bar{\text{b}} - \text{y}\bar{\text{c}}) \).
Because \( \bar{\text{a}} \) is perpendicular to others, cross terms with \( \bar{\text{a}} \) vanish:
\( |\bar{\text{d}}|^2 = |\bar{\text{a}}|^2 + |(1 + 2\text{y})\bar{\text{b}} - \text{y}\bar{\text{c}}|^2 \)
\( = 1 + (1 + 2\text{y})^2|\bar{\text{b}}|^2 + \text{y}^2|\bar{\text{c}}|^2 - 2\text{y}(1 + 2\text{y})(\bar{\text{b}} \cdot \bar{\text{c}}) \)
\( = 1 + (1 + 4\text{y} + 4\text{y}^2)(4) + \text{y}^2(1) - 2\text{y}(1 + 2\text{y})(1) \)
\( = 1 + 4 + 16\text{y} + 16\text{y}^2 + \text{y}^2 - 2\text{y} - 4\text{y}^2 \)
\( = 5 + 14\text{y} + 13\text{y}^2 \).
Step 4: Final Answer:
The value of \( |\bar{\text{d}}|^2 \) is \( 13\text{y}^2 + 14\text{y} + 5 \).