Question

A vector $\vec{n}$ is inclined to X-axis at 45°, Y-axis at 60° and at an acute angle to Z-axis. If $\vec{n}$ is normal to a plane passing through the point $(-\sqrt{2},1,1)$ then equation of the plane is

• A. $\sqrt{2}x+y+z=0$
• B. $x+\sqrt{2}y+z=1$
• C. $-\sqrt{2}x+y+2z=5$
• D. $x+y+\sqrt{2}z=1$

Hint:

Logic Tip: As soon as you discover the normal vector's direction ratios are $(\sqrt{2}, 1, 1)$, check the options. Option A is the only one with this sequence of coefficients! You don't even need to plug in the point.

Answer 1

The Correct Option is A

Concept:
If a vector makes angles $\alpha, \beta, \gamma$ with the coordinate axes, its direction cosines satisfy the identity $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. The normal vector can be formed using these direction cosines. The equation of the plane is then $l(x-x_1) + m(y-y_1) + n(z-z_1) = 0$.
Step 1: Find the direction cosine for the Z-axis ($\gamma$).
Given angles are $\alpha = 45^\circ$, $\beta = 60^\circ$, and $\gamma$ is acute. Using the direction cosines property: $$\cos^2(45^\circ) + \cos^2(60^\circ) + \cos^2\gamma = 1$$ $$\left(\frac{1}{\sqrt{2\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2\gamma = 1$$ $$\frac{1}{2} + \frac{1}{4} + \cos^2\gamma = 1$$ $$\frac{3}{4} + \cos^2\gamma = 1 \implies \cos^2\gamma = \frac{1}{4}$$ Since $\gamma$ is an acute angle, $\cos\gamma$ must be positive: $$\cos\gamma = \frac{1}{2} \implies \gamma = 60^\circ$$
Step 2: Write the normal vector $\vec{n}$.
The direction cosines $(l, m, n)$ are $(\frac{1}{\sqrt{2, \frac{1}{2}, \frac{1}{2})$. The normal vector is: $$\vec{n} = \frac{1}{\sqrt{2\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$$ To simplify, we can multiply the direction ratios by a scalar (like 2) without changing the normal's direction: $$2\vec{n} = \sqrt{2}\hat{i} + 1\hat{j} + 1\hat{k}$$ So, use direction ratios $(\sqrt{2}, 1, 1)$ for the plane's normal.
Step 3: Find the Cartesian equation of the plane.
The plane passes through the point $(-\sqrt{2}, 1, 1)$. Using the point-normal form: $$\sqrt{2}(x - (-\sqrt{2})) + 1(y - 1) + 1(z - 1) = 0$$ $$\sqrt{2}(x + \sqrt{2}) + y - 1 + z - 1 = 0$$ $$\sqrt{2}x + 2 + y + z - 2 = 0$$ $$\sqrt{2}x + y + z = 0$$