Consider the vectors a = 2 i + 3 j + 4 k and b = 3 i + 4 j + 5 k ....

The given vectors are $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} and$\vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.<br><br>The dot product of the vectors is:<br>$\vec{a} \cdot \vec{b} = (2)(3) + (3)(4) + (4)(5) = 6 + 12 + 20 = 38$<br><br>The magnitudes of the vectors are:<br>$|\vec{a}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$<br>$|\vec{b}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$<br><br>The angle$\theta$between the vectors is given by:<br>$\cos \theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$<br><br>Substituting the values:<br>$\cos \theta = \dfrac{38}{\sqrt{29} \sqrt{50}} = \dfrac{38}{\sqrt{1450}}$<br><br>$\theta = \cos^{-1} $\left(\dfrac{38}{\sqrt{1450}}\right)$