The plane 2x – 3y + 6z – 11 = 0 makes an angle sin^–1(α) with x-axis. The value of α is equal to

Concept:

If there exists a line \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\), having direction ratios (a, b, c) and a plane Ax + By + Cz + D = 0 having direction ratios (A, B, C) then the acute angle α between the line and the plane is given by:

sin θ = \(\left|\rm \frac{aA+bB+cC}{\sqrt{a^2+b^2+c^2}\sqrt{A^2+B^2+C^2}}\right|\)

Calculation:

Given: The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis

∴ Direction ratios of the plane = (A, B, C) = (2, -3, 6)

and, direction ratios of x-axis = (a, b, c) = (1, 0, 0)

∴ sin θ = \(\left|\rm \frac{aA+bB+cC}{\sqrt{a^2+b^2+c^2}\sqrt{A^2+B^2+C^2}}\right|\)

= \(\left|\rm \frac{1\times2+0\times(-3)+0\times6}{\sqrt{1^2+0^2+0^2}\sqrt{2^2+(-3)^2+6^2}}\right|\)

= \(\left|\rm \frac{2+0+0}{\sqrt{1+0+0}\sqrt{4+9+36}}\right|\)

= \(\frac{2}{\sqrt{49}}\)

= \(\frac{2}{7}\)

⇒ sin θ = \(\frac{2}{7}\)

⇒ θ = sin–1(\(\frac{2}{7}\))

∴ α = \(\frac{2}{7}\)

The value of α is equal to 2/7.

The correct answer is option 3.