Question

Find the vector and Cartesian equation of the plane which passes through the point $(1,0,-2)$ and on which the vector $\hat{i} + \hat{j} - \hat{k}$ is perpendicular.

Hint:

For a plane equation, use $\vec{n}\cdot(\vec{r}-\vec{r}_0)=0$, where $\vec{n}$ is the normal vector.

Answer 1

Step 1: Recall formula. The equation of a plane passing through point $\vec{r}_0$ with normal vector $\vec{n}$ is: \[ \vec{n} \cdot (\vec{r} - \vec{r}_0) = 0 \]

Step 2: Write given data. Normal vector: $\vec{n} = \hat{i} + \hat{j} - \hat{k} = (1,1,-1)$ Point: $(1,0,-2)$

Step 3: Vector equation of plane. \[ (1,1,-1) \cdot \big( (x,y,z) - (1,0,-2) \big) = 0 \] \[ (x-1) + (y-0) - (z+2) = 0 \] \[ x + y - z - 3 = 0 \]

Step 4: Write both forms. - Vector form: \[ \vec{r} \cdot (1,1,-1) = (1,0,-2)\cdot(1,1,-1) \] \[ \vec{r} \cdot (1,1,-1) = 3 \] - Cartesian form: \[ x + y - z - 3 = 0 \]

Final Answer: Vector equation: $\;\boxed{\vec{r}\cdot(1,1,-1)=3}$ Cartesian equation: $\;\boxed{x+y-z-3=0}$