Question

Let \[ \vec a=\hat i+2\hat j+2\hat k, \qquad \vec b=2\hat i+\hat j+2\hat k. \] If \(\theta\) is the angle between \(\vec a\) and \(\vec b\), then the value of \[ \cos\theta \] is:

• A. \(\frac{8}{9}\)
• B. \(\frac{7}{9}\)
• C. \(\frac{5}{9}\)
• D. \(\frac{4}{9}\)

Hint:

For vectors \[ (a,b,c) \quad\text{and}\quad (p,q,r), \] always remember \[ \cos\theta = \frac{ap+bq+cr} {\sqrt{a^2+b^2+c^2}\sqrt{p^2+q^2+r^2}}. \] This is the standard formula for finding the angle between two vectors.

Answer 1

The Correct Option is A

Concept: The angle between two vectors is determined using the dot product formula \[ \vec a\cdot\vec b = |\vec a|\,|\vec b|\cos\theta. \] From this relation, \[ \cos\theta = \frac{\vec a\cdot\vec b} {|\vec a||\vec b|}. \] This formula converts a geometric angle problem into a straightforward algebraic computation involving vector components.

Step 1: Write the vectors in component form. Given, \[ \vec a=(1,2,2), \] and \[ \vec b=(2,1,2). \]

Step 2: Calculate the dot product. Using \[ \vec a\cdot\vec b = a_1b_1+a_2b_2+a_3b_3, \] we obtain \[ \vec a\cdot\vec b = (1)(2)+(2)(1)+(2)(2). \] \[ = 2+2+4. \] \[ =8. \]

Step 3: Find the magnitude of \(\vec a\). \[ |\vec a| = \sqrt{1^2+2^2+2^2}. \] \[ = \sqrt{1+4+4}. \] \[ = \sqrt9. \] \[ =3. \]

Step 4: Find the magnitude of \(\vec b\). \[ |\vec b| = \sqrt{2^2+1^2+2^2}. \] \[ = \sqrt{4+1+4}. \] \[ = \sqrt9. \] \[ =3. \]

Step 5: Apply the angle formula. \[ \cos\theta = \frac{8}{3\times3}. \] \[ = \frac89. \]

Step 6: Verification. Since \[ 0<\frac89<1, \] the value is valid for a cosine and indicates that the angle between the vectors is acute.

Step 7: Final Conclusion. \[ \boxed{\cos\theta=\frac89} \] Hence the correct answer is \[ \boxed{\text{Option (A)}}. \]