Question

Let \(\overrightarrow a\) and \(\overrightarrow b\) be two unit vectors. If the vectors \(\overrightarrow c=5\overrightarrow a-4\overrightarrow b\) and \(\overrightarrow d = \overrightarrow a+2\overrightarrow b\) perpendicular to each other, then the angle between \(\overrightarrow a\) and \(\overrightarrow b\) is:

• A. 0
• B. \(\frac{π}{3}\)
• C. \(\frac{π}{4}\)
• D. \(\frac{π}{6}\)

Answer 1

The Correct Option is B

Given that the vectors \(\overrightarrow c=5\overrightarrow a-4\overrightarrow b\) and \(\overrightarrow d = \overrightarrow a+2\overrightarrow b\) are perpendicular, we know their dot product is zero. Therefore, we have:

\( (5\overrightarrow a - 4\overrightarrow b) \cdot (\overrightarrow a + 2\overrightarrow b) = 0 \)

Expanding this, we get:

\( 5(\overrightarrow a \cdot \overrightarrow a) + 10(\overrightarrow a \cdot \overrightarrow b) - 4(\overrightarrow b \cdot \overrightarrow a) - 8(\overrightarrow b \cdot \overrightarrow b) = 0 \)

Since \(\overrightarrow a\) and \(\overrightarrow b\) are unit vectors, we have \(\overrightarrow a \cdot \overrightarrow a = 1\) and \(\overrightarrow b \cdot \overrightarrow b = 1\). Therefore, the equation simplifies to:

\( 5 + (10 - 4)(\overrightarrow a \cdot \overrightarrow b) - 8 = 0 \)

Simplify further:

\( 5 + 6(\overrightarrow a \cdot \overrightarrow b) - 8 = 0 \)

\( 6(\overrightarrow a \cdot \overrightarrow b) = 3 \)

\(\overrightarrow a \cdot \overrightarrow b = \frac{1}{2} \)

Thus, the cosine of the angle \(\theta\) between \(\overrightarrow a\) and \(\overrightarrow b\) is \(\frac{1}{2}\), which corresponds to an angle of:

\(\theta = \frac{\pi}{3} \)

Therefore, the angle between \(\overrightarrow a\) and \(\overrightarrow b\) is: \(\frac{\pi}{3}\)