Find the angle between the body diagonals of a cube

Griffiths 1

Pairoaj

From the figure,

$$A = (1-0)\hat{x}+(1-0)\hat{y}+(0-1)\hat{z}=1\hat{x}+1\hat{y}-1\hat{z}$$ $$\left\lvert A \right\rvert=\sqrt{1^2+1^2+(-1)^2}=\sqrt{3}$$

$$B = (1-0)\hat{x}+(1-0)\hat{y}+(1-0)\hat{z}=1\hat{x}+1\hat{y}+1\hat{z}$$ $$\left\lvert B \right\rvert=\sqrt{1^2+1^2+1^2}=\sqrt{3}$$

now use the relationship

$$A \cdot B=AB \cos{\theta}$$

$$\rightarrow (1,1,-1) \cdot(1,1,1)=\sqrt{3} \sqrt{3} \cos{\theta}$$

$$\rightarrow 1=3 \cos{\theta}$$

$$\rightarrow \cos{\theta}=\frac{1}{3}$$

$$\rightarrow \theta=\cos^{-1}{\frac{1}{3}}\approx70.5^\circ$$

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Asked: 2 years ago