Question

Let a, b, x, y be real numbers such that \(a^2+b^2=25,x^2+y^2=169,\) and \(ax+by=65.\) If \(k= ay-bx,\) then

• A. k=0
• B. \(0<k≤\frac{5}{13}\)
• C. \(k=\frac{5}{13}\)
• D. \(k>\frac{5}{13}\)

Answer 1

The Correct Option is A

Given: 

• \( a^2 + b^2 = 25 \)
• \( x^2 + y^2 = 169 \)
• \( ax + by = 65 \)

Required:

Find the value of \( k = ay - bx \).

Step 1: Vector Interpretation

Let \( \vec{a} = (a, b) \), and \( \vec{x} = (x, y) \). Then:

• \( |\vec{a}| = \sqrt{a^2 + b^2} = \sqrt{25} = 5 \)
• \( |\vec{x}| = \sqrt{x^2 + y^2} = \sqrt{169} = 13 \)
• Dot product: \( \vec{a} \cdot \vec{x} = ax + by = 65 \)

Using the dot product formula:

\[ \vec{a} \cdot \vec{x} = |\vec{a}||\vec{x}|\cos\theta = 5 \times 13 \times \cos\theta = 65 \] \[ \Rightarrow \cos\theta = \frac{65}{65} = 1 \Rightarrow \theta = 0^\circ \]

This means \( \vec{a} \) and \( \vec{x} \) are in the same direction (parallel).

Step 2: Proportional Vectors

Since the vectors are parallel, we can write:

\[ (a, b) = m(x, y) \Rightarrow a = mx,\quad b = my \]

Step 3: Evaluate \( k = ay - bx \)

Substituting:

\[ k = ay - bx = (my)y - (mx)x = m(y^2 - x^2) \]

Step 4: Use the fact that \( \vec{a} \parallel \vec{x} \)

Since the vectors are proportional, \( y = \pm x \Rightarrow y^2 = x^2 \)

\[ \Rightarrow y^2 - x^2 = 0 \Rightarrow k = m \cdot 0 = 0 \]

Final Answer:

\( \boxed{k = 0} \)