Question

The vector \(\overset{̄}{a}+3\overset{̄}{b}\) is perpendicular to \(7\overset{̄}{a}-5\overset{̄}{b}\) and the vector \(\overset{̄}{a}-4\overset{̄}{b}\) is perpendicular to \(7\overset{̄}{a}-2\overset{̄}{b}\). Then the angle between \(\overset{̄}{a}\) and \(\overset{̄}{b}\) is

• A. \(\frac{π}{2}\)
• B. \(\frac{π}{4}\)
• C. \(\frac{π}{6}\)
• D. \(\frac{π}{3}\)

Hint:

Use dot product zero for each perpendicular pair.

Answer 1

The Correct Option is D

Step 1: Expand the first condition
\((\vec a+3\vec b)\cdot(7\vec a-5\vec b)=0\) gives \(7|a|^2+16\,\vec a\cdot\vec b-15|b|^2=0\).

Step 2: Expand the second condition
\((\vec a-4\vec b)\cdot(7\vec a-2\vec b)=0\) gives \(7|a|^2-30\,\vec a\cdot\vec b+8|b|^2=0\).

Step 3: Solve
Subtract: \(46\,\vec a\cdot\vec b-23|b|^2=0\), so \(\vec a\cdot\vec b=\frac{|b|^2}{2}\).
Substitute into the second equation: \(7|a|^2-15|b|^2+8|b|^2=0\), so \(|a|=|b|\).

Step 4: Angle
\(\cos\theta=\frac{|b|^2/2}{|a||b|}=\frac12\), so \(\theta=\frac{\pi}{3}\).

Final Answer:
The angle is \(\frac{\pi}{3}\), option (D). \[ \boxed{\text{(D) } \frac{\pi}{3}} \]