Question

The angle at which the normal to the plane 4x - 8y + z = 7 is inclined to y-axis is :

• A. \(\cos^{-1}(\frac{4}{9})\)
• B. \(\cos^{-1}(\frac{1}{9})\)
• C. \(\cos^{-1}(\frac{8}{9})\)
• D. \(\cos^{-1}(\frac{5}{9})\)

Answer 1

The Correct Option is C

To determine the angle at which the normal to the plane \(4x - 8y + z = 7\) is inclined to the y-axis, we first need to identify the normal vector to the plane. The plane equation is of the form \(Ax + By + Cz = D\). Thus, the normal vector \(\vec{N}\) can be derived as \(\vec{N} = \langle 4, -8, 1 \rangle\).

The angle \(\theta\) between the normal vector \(\vec{N}\) and the y-axis vector \(\vec{j} = \langle 0, 1, 0 \rangle\) can be found using the dot product formula:

\[\cos \theta = \frac{\vec{N} \cdot \vec{j}}{|\vec{N}| \cdot |\vec{j}|}\]

Calculating the dot product:

\(\vec{N} \cdot \vec{j} = 4 \cdot 0 + (-8) \cdot 1 + 1 \cdot 0 = -8\)

The magnitude of \(\vec{N}\) is:

\(|\vec{N}| = \sqrt{4^2 + (-8)^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9\)

The magnitude of \(\vec{j}\) is:

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

Substituting these into the cosine formula:

\[\cos \theta = \frac{-8}{9 \cdot 1} = -\frac{8}{9}\]

The angle between the y-axis and the normal vector, considering the magnitude, is \(\theta = \cos^{-1}\left(-\frac{8}{9}\right)\), but since angles made by normal are conventionally taken as positive, we determine the complementary angle \(\cos^{-1}\left(\frac{8}{9}\right)\).

Thus, the correct angle is \(\boxed{\cos^{-1}\left(\frac{8}{9}\right)}\).