Question

Find the angle between the lines \(\vec r = 3\hat i+2\hat j-4\hat k+\lambda(\hat i+2\hat j+2\hat k)\) and \(\vec r = 5\hat i-2\hat j+\mu(3\hat i+2\hat j+6\hat k)\).

Hint:

Use cosθ = |d1·d2|/(|d1||d2|) with the direction vectors of both lines.

Answer 1

Step 1: Pull out the direction vectors and normalise each into a unit vector first:
$\vec d_1=(1,2,2)$ has magnitude $\sqrt{1+4+4}=3$, so its unit vector is $\left(\dfrac13,\dfrac23,\dfrac23\right)$. $\vec d_2=(3,2,6)$ has magnitude $\sqrt{9+4+36}=7$, so its unit vector is $\left(\dfrac37,\dfrac27,\dfrac67\right)$.

Step 2: Dot the two unit vectors, which directly gives $\cos\theta$:
$\left(\dfrac13\right)\left(\dfrac37\right)+\left(\dfrac23\right)\left(\dfrac27\right)+\left(\dfrac23\right)\left(\dfrac67\right) = \dfrac{3}{21}+\dfrac{4}{21}+\dfrac{12}{21}=\dfrac{19}{21}$.

Step 3: State the angle:
$\cos\theta=\dfrac{19}{21}$, so $\theta=\cos^{-1}\left(\dfrac{19}{21}\right)$.

Final Answer:
The angle between the two lines is $\cos^{-1}(19/21)$. \[ \boxed{\theta=\cos^{-1}\left(\dfrac{19}{21}\right)} \]