The angle between the lines r=3 i+2 j-4 k+λ(i+2 j+2 k) and r=5 j-2 k+…

$\vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})$ → line 1 (l₁)

$\vec{r}=5 \hat{j}-2 \hat{k}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})$ → line 2 (l₂)

vector

$(\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}) || l_1$

$(\vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}) || l_2$

So angle between $l_1$ and $l_2=$ angle between $\vec{a}$ aud $\vec{b}$ so $\vec{a} . \vec{b}=|\vec{a}||\vec{b}| \cos \theta$

$=(\hat{i}+2 \hat{j}+2 \hat{k})(3 \hat{i}+2 \hat{j}+6 \hat{k})=\sqrt{1+4+4} \sqrt{9+4+36} \cos \theta$

$\Rightarrow \frac{3+4+12}{\sqrt{9} \sqrt{49}}=\cos \theta$

$\Rightarrow \cos \theta=\frac{19}{3 \times 7}=\frac{19}{21}$

$\Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)$