The angle between a normal to the plane 2x - | vedclass

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Question

(A) The equation of the plane is given by $2x - y + 2z - 1 = 0$. The normal vector $\vec{n}$ to this plane is given by the coefficients of $x, y,$ and

Vedclass · Accepted Answer

$\cos^{-1} \frac{2}{3}$

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(A) The equation of the plane is given by $2x - y + 2z - 1 = 0$. The normal vector $\vec{n}$ to this plane is given by the coefficients of $x, y,$ and $z$, which is $\vec{n} = 2\hat{i} - \hat{j} + 2\hat{k}$. The direction vector of the $X$-axis is $\vec{a} = 1\hat{i} + 0\hat{j} + 0\hat{k}$. The angle $	heta$ between the normal vector $\vec{n}$ and the $X$-axis is given by the formula $\cos 	heta = \frac{|\vec{n} \cdot \vec{a}|}{|\vec{n}| |\vec{a}|}$. Calculating the dot product: $\vec{n} \cdot \vec{a} = (2)(1) + (-1)(0) + (2)(0) = 2$. Calculating the magnitudes: $|\vec{n}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ and $|\vec{a}| = \sqrt{1^2 + 0^2 + 0^2} = 1$. Therefore, $\cos 	heta = \frac{2}{3 	imes 1} = \frac{2}{3}$. Thus, $	heta = \cos^{-1} \frac{2}{3}$.

The angle between a normal to the plane $2x - y + 2z - 1 = 0$ and the $X$-axis is

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