Question

Find the angle between the two vectors A = 2i + 3j -k and B = i -2j + 3k.​

Answer 1

ATQ:

$\sf \overrightarrow{\sf A} = 2\hat{i} + 3\hat{j} - \hat{k}$

$\sf \overrightarrow{\sf B} = \hat{i} - 2\hat{j} + 3\hat{k}$

To find:

The angle between the two vectors A & B.

Solution:

We know that,

$\sf \Longrightarrow cos \theta = \dfrac{\overrightarrow{\sf A}.\overrightarrow{\sf B}}{|\overrightarrow{\sf A}| |\overrightarrow{\sf B}|}$

$\sf \Longrightarrow cos \theta = \dfrac{\Big(2\hat{i} + 3\hat{j} - \hat{k} \Big) . \Big(\hat{i} - 2\hat{j} + 3\hat{k} \Big)}{\Big(\sqrt{(2)^2 + (3)^2 + (-1)^2}\Big)\Big(\sqrt{(1)^2 + (-2)^2 + (3)^2}\Big)}$

We also know that:

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

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

$\sf \Longrightarrow cos \theta = \dfrac{(2\hat{i})(1\hat{i}) + (3\hat{j})(-2\hat{j}) -(1\hat{k})(3\hat{k})}{\Big(\sqrt{4 + 9 + 1}\Big)\Big(\sqrt{1 + 4 + 9}\Big)}$

$\sf \Longrightarrow cos \theta = \dfrac{2 - 6 - 3}{\Big(\sqrt{14}\Big)\Big(\sqrt{14}\Big)}$

$\sf \Longrightarrow cos \theta = \dfrac{2 - 9}{14}$

$\sf \Longrightarrow cos \theta = \dfrac{-7}{14}$

$\sf \Longrightarrow cos \theta = \dfrac{-1}{2}$

$\sf \Longrightarrow \theta = cos^{-1} \ \Bigg\{\dfrac{-1}{2}\Bigg\}$

$\sf \Longrightarrow \theta = 120^\circ$

Therefore, the angle between the two vectors $\sf \overrightarrow{\sf A} \ and \ \overrightarrow{\sf B}$ is 120°.

Answer 2

Answer:

The angle between two vectors a and b is 120°

hope this helps you ☺️