Question

The angle between the vector i-j and j-k is

Answer 1

The angle between the vector i-j and j-k is  $\dfrac{2\pi}{3}$ .

Explanation:

We know that the formula to find the angle between two vectors u and v is given by :-

$\cos\theta=\dfrac{u\cdot v}{|u||v|}$

Let,

$u=\hat{i}-\hat{j}\\\\ v=\hat{j}-\hat{k}$

Then,

$u\cdot v=(\hat{i}-\hat{j})\cdot (\hat{j}-\hat{k})=\hat{i}\cdot 0+(\hat{-j})\cdot\hat{j}+0\cdot \hat{-k}\\\\=-\hat{j}^2=-1$

Also,

$|u|=|i-j|=\sqrt{1+1}=\sqrt{2}\\\\ |v|=|j-k|=\sqrt{1+1}=\sqrt{2}$

Substitute values in formula , we get

$\cos\theta=\dfrac{-1}{(\sqrt{2})(\sqrt{2})}\\\\ \cos\theta=\dfrac{-1}{2}\\\\ \theta=\cos^{-1}(\dfrac{-1}{2})\\\\ \theta = \dfrac{2\pi}{3}$

Hence, the angle between the vector i-j and j-k is  $\dfrac{2\pi}{3}$ .

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