Question

Let a = 2i+3j+4k and b = 3i+4j+5k. Find the angle between them

Answer 1

it just easy just apply the formulas a.bvector/a×bvector

Answer 2

Explanation:

We have,

$\rm \vec{a} = 2\hat{i} + 3 \hat{j} + 4 \hat{k}$

$\rm \vec{b} = 3\hat{i} + 4 \hat{j} + 5 \hat{k}$

We know that,

$\rm \cos θ = \dfrac{\vec{a} . \vec{b} }{\mid \vec{a} \mid \mid \vec{b} \mid }$

$\rm \cos θ = \dfrac{(2 \hat{i} +3 \hat{j} + 4 \hat{k})(3 \hat{i} +4 \hat{j} +5 \hat{k})}{\sqrt{2^2+3^2+4^2}\sqrt{3^2+4^2+5^2}}$

$\rm \cos θ = \dfrac{3×2+3×4+4×5}{\sqrt{29}{\sqrt{50}}}$

$\rm \cos θ = \dfrac{6+12+20}{\sqrt{29}{\sqrt{50}}}$

$\rm \cos θ = \dfrac{38}{\sqrt{1450}}$

$\rm θ = \cos ^{-1} \bigg( \dfrac{38}{\sqrt{1450}} \bigg)$