Question

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

• A. \(cos^{-1}\frac{2}{3}\)
• B. \(cos^{-1}\frac{1}{5}\)
• C. \(cos^{-1}\frac{3}{4}\)
• D. \(cos^{-1}\frac{1}{3}\)

Answer 1

The Correct Option is A

Given the equation of the plane \(2x - y + 2z - 1 = 0\), the normal vector to the plane is given by the coefficients of x, y, and z:

\(\vec{n} = (2, -1, 2)\)

The direction vector of the X-axis is given by: 

\(\vec{i} = (1, 0, 0)\)

The angle \(\theta\) between the normal vector and the X-axis can be found using the dot product formula:

\(\vec{n} \cdot \vec{i} = ||\vec{n}|| \cdot ||\vec{i}|| \cdot \cos{\theta}\)

The dot product is:

\(\vec{n} \cdot \vec{i} = (2)(1) + (-1)(0) + (2)(0) = 2\)

The magnitudes are:

\(||\vec{n}|| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)

\(||\vec{i}|| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1\)

Now, we can find the cosine of the angle:

\(\cos{\theta} = \frac{\vec{n} \cdot \vec{i}}{||\vec{n}|| \cdot ||\vec{i}||} = \frac{2}{3 \cdot 1} = \frac{2}{3}\)

Therefore, the angle \(\theta\) is:

\(\theta = \cos^{-1}\left(\frac{2}{3}\right)\)

Final Answer: \(\cos^{-1}\left(\frac{2}{3}\right)\)