Question

A plane passes through the point $(0,1,1)$ and has normal vector $\hat{i}+\hat{j}+\hat{k}$. Its equation is

• A. $x+y+z=1$
• B. $x+y+z=2$
• C. $x+2y+2z=1$
• D. $y+z=2$
• E. $y+z=1$

Hint:

Geometry Tip: If the normal vector is $\hat{i}+\hat{j}+\hat{k}$, the equation must be of the form $x+y+z = d$. You can find '$d$' instantly by plugging the given point $(0,1,1)$ into $x+y+z$, getting $0+1+1 = 2$.

Answer 1

The Correct Option is B

Concept:
The Cartesian equation of a plane passing through a specific point $(x_1, y_1, z_1)$ with a normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by the formula $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.

Step 1: Identify the normal vector components.
The normal vector is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$. The scalar components $(a, b, c)$ of the normal vector are $(1, 1, 1)$.

Step 2: Identify the given point on the plane.
The plane passes through the point $(x_1, y_1, z_1) = (0, 1, 1)$.

Step 3: Substitute values into the plane equation.
Using the formula $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$: $$1(x - 0) + 1(y - 1) + 1(z - 1) = 0$$

Step 4: Expand the algebraic equation.
Distribute the coefficients (which are just 1) into the parentheses: $$x + y - 1 + z - 1 = 0$$

Step 5: Simplify to match the standard form.
Combine the constant terms and move them to the right side of the equation: $$x + y + z - 2 = 0$$ $$x + y + z = 2$$ Hence the correct answer is (B) $x+y+z=2$.