Question

Find the angle between vectors \(\hat i-2\hat j+3\hat k\) and \(3\hat i-2\hat j+\hat k\).

Hint:

Use cos(theta) = (a.b)/(|a||b|) with the dot product and magnitudes of the two vectors.

Answer 1

Step 1: Compute the dot product:
\[ (1)(3)+(-2)(-2)+(3)(1)=3+4+3=10 \]

Step 2: Compute the magnitudes:
\[ |\vec a|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14},\quad |\vec b|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{14} \]

Step 3: Use \(\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}\):
\[ \cos\theta=\dfrac{10}{\sqrt{14}\cdot\sqrt{14}}=\dfrac{10}{14}=\dfrac57 \]

Final Answer:
\[ \boxed{\theta=\cos^{-1}\left(\dfrac57\right)} \]