Question

if a=2i^ +3j^+4k^ and +4 i^ 3j^+2k^ find the angle between a and b​

Answer 1

ATQ:

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

$\rm \overrightarrow{\rm b} = 4\hat{i} + 3\hat{j} + 2\hat{k}$

To find:

Angle between $\overrightarrow{\rm a}$ and $\overrightarrow{\rm b}$.

Solution:

We know that:

$\sf \Longrightarrow cos \theta = \dfrac{\overrightarrow{\sf a} . \overrightarrow{\sf b}}{|\overrightarrow{\sf a}| \ |\overrightarrow{\sf b}|}$

$\sf \Longrightarrow cos \theta = \dfrac{\Big(2\hat{i} + 3\hat{j} + 4\hat{k}\Big)\Big(4\hat{i} + 3\hat{j} + 2\hat{k}\Big)}{\Big(\sqrt{(2)^2 +(3)^2 + (4)^2}\Big) \ \Big(\sqrt{(4)^2 + (3)^2 + (2)^2}\Big)}$

We know that:

$\dashrightarrow \ \hat{\sf i} \times \hat{\sf i} = \hat{\sf j} \times \hat{\sf j} = \hat{\sf k} \times \hat{\sf k} = 1$

$\dashrightarrow \ \hat{\sf i} \times \hat{\sf j} = \hat{\sf j} \times \hat{\sf k} = \hat{\sf k} \times \hat{\sf i} = 0$

Therefore:

$\sf \Longrightarrow cos \theta = \dfrac{(2\hat{i})(4\hat{i}) + (3\hat{j})(3\hat{j}) + (4\hat{k})(2\hat{k})}{\Big(\sqrt{4 + 9 + 16}\Big) \ \Big(\sqrt{16 + 9 + 4}\Big)}$

$\sf \Longrightarrow cos \theta = \dfrac{8 + 9 + 8}{\sqrt{29} \ \times \sqrt{29}}$

$\sf \Longrightarrow cos \theta = \dfrac{25}{29}$

$\sf \Longrightarrow \theta = cos^{-1} \ \dfrac{25}{29}$

Hence the angle between $\overrightarrow{\sf a} \sf and \overrightarrow{\sf b}$ is cos⁻¹ (25/29).