Question

Find the vector equation of the plane passing through the point having position vector \( 2\hat{i} + 3\hat{j} + 4\hat{k} \) and perpendicular to the vector \( 2\hat{i} + \hat{j} - 2\hat{k} \).

Hint:

The equation of a plane can be written in vector form as \( \vec{r} \cdot \vec{n} = \vec{r_0} \cdot \vec{n} \), where \( \vec{n} \) is the normal vector and \( \vec{r_0} \) is a point on the plane.

Answer 1

Step 1: Recall the equation of a plane.
The vector equation of a plane passing through a point \( \vec{r_0} = x_0 \hat{i} + y_0 \hat{j} + z_0 \hat{k} \) and perpendicular to a normal vector \( \vec{n} = a \hat{i} + b \hat{j} + c \hat{k} \) is given by: \[ \vec{r} \cdot \vec{n} = \vec{r_0} \cdot \vec{n} \]

Step 2: Identify the given quantities.
The position vector of the point is \( \vec{r_0} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and the normal vector to the plane is \( \vec{n} = 2\hat{i} + \hat{j} - 2\hat{k} \).

Step 3: Write the equation of the plane.
The vector equation of the plane is: \[ \vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (2\hat{i} + \hat{j} - 2\hat{k}) \]

Step 4: Compute the dot product on the right-hand side.
\[ (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 2 \times 2 + 3 \times 1 + 4 \times (-2) = 4 + 3 - 8 = -1 \]

Step 5: Final equation of the plane.
Thus, the equation of the plane is: \[ \vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = -1 \]

Final Answer: \[ \boxed{\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = -1} \]