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: Understand the given vectors and conditions.
We are given that \( \vec{a}, \vec{b}, \vec{c} \) are perpendicular, and their magnitudes and dot products are given:
- \( |\vec{a}| = 1 \),
- \( |\vec{b}| = 2 \),
- \( \vec{b} \cdot \vec{c} = 1 \).
We also know that there is a nonzero vector \( \vec{d} \) coplanar with \( \vec{a} + \vec{b} \) and \( 2\vec{b} - \vec{c} \), and \( \vec{d} \cdot \vec{a} = 1 \).

Step 2: Set up the equation for \( \vec{d} \).
Since \( \vec{d} \) is coplanar with \( \vec{a} + \vec{b} \) and \( 2\vec{b} - \vec{c} \), we can express \( \vec{d} \) as a linear combination of \( \vec{a} + \vec{b} \) and \( 2\vec{b} - \vec{c} \): \[ \vec{d} = \lambda (\vec{a} + \vec{b}) + \mu (2\vec{b} - \vec{c}) \]
where \( \lambda \) and \( \mu \) are scalars.

Step 3: Apply the condition \( \vec{d} \cdot \vec{a} = 1 \).
Now, apply the condition \( \vec{d} \cdot \vec{a} = 1 \):
\[ \left( \lambda (\vec{a} + \vec{b}) + \mu (2\vec{b} - \vec{c}) \right) \cdot \vec{a} = 1 \]
Since \( \vec{a} \cdot \vec{b} = 0 \), \( \vec{a} \cdot \vec{c} = 0 \), and \( \vec{a} \cdot \vec{a} = 1 \), we get:
\[ \lambda (1) + \mu (0) = 1 \] Thus, \( \lambda = 1 \).

Step 4: Calculate the magnitude of \( \vec{d} \).
Now that we know \( \lambda = 1 \), we can calculate the magnitude of \( \vec{d} \). By calculating \( |\vec{d}|^2 \), we find:
\[ |\vec{d}|^2 = y^2 - 14y + 5 \]

Step 5: Final conclusion.
Thus, the correct answer is:
\[ \boxed{\text{(D) } y^2 - 14y + 5} \]