Question

The distance of the point \( O(0,0,0) \) from the plane \( \overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 5 \) measured parallel to \( 2\hat{i} + 3\hat{j} - 6\hat{k} \) is?

• A. \( 35 \)
• B. \( 30 \)
• C. \( 25 \)
• D. \( 42 \)

Hint:

When measuring distance from a point to a plane along a given direction, parameterize the line and solve for \( \lambda \) using the plane equation. The final distance is obtained by scaling the magnitude of the direction vector.

Answer 1

The Correct Option is A

We are given the equation of the plane: \[ \overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 5 \] and we need to find the distance of the point \( O(0,0,0) \) from the plane, measured parallel to the direction \( 2\hat{i} + 3\hat{j} - 6\hat{k} \). Step 1: Find the normal to the plane
The normal to the given plane is: \[ \overrightarrow{N} = \hat{i} + \hat{j} + \hat{k}. \] Step 2: Find the equation of the line
The given direction \( \overrightarrow{D} = 2\hat{i} + 3\hat{j} - 6\hat{k} \) represents the direction along which we measure the distance. A general point on this line through \( O(0,0,0) \) is given by: \[ \overrightarrow{r} = \lambda (2\hat{i} + 3\hat{j} - 6\hat{k}). \] Step 3: Find intersection of line with plane
Substituting this point into the equation of the plane: \[ (2\lambda) + (3\lambda) + (-6\lambda) = 5 \] \[ 2\lambda + 3\lambda - 6\lambda = 5 \] \[ - \lambda = 5 \quad \Rightarrow \quad \lambda = -5. \] Step 4: Find the required distance
Since the displacement along the given direction is \( \lambda \) times the magnitude of the direction vector, \[ \text{Distance} = |\lambda| \times |\overrightarrow{D}|. \] \[ |\overrightarrow{D}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7. \] Thus, the required distance is: \[ |-5| \times 7 = 35. \]

Answer 2

Step 1: Convert plane to scalar form
We are given the vector form of the plane: \[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 5 \Rightarrow x + y + z = 5 \] This is the scalar form of the plane equation.

Step 2: Take any point on the plane
Let us choose a simple point on the plane. Suppose we take: \[ x = 5,\ y = 0,\ z = 0 \Rightarrow (5, 0, 0) \] This point lies on the plane because \(5 + 0 + 0 = 5\).

Step 3: Form vector from origin to the plane
Now form a vector from the origin \(O(0, 0, 0)\) to the point on the plane: \[ \vec{A} = \overrightarrow{OA} = 5\hat{i} \]

Step 4: Project this vector onto given direction
We are to find the distance from the origin to the plane measured **parallel to** the vector: \[ \vec{D} = 2\hat{i} + 3\hat{j} - 6\hat{k} \] We use projection of \(\vec{A}\) on \(\vec{D}\): \[ \text{Distance} = \left| \frac{\vec{A} \cdot \vec{D}}{|\vec{D}|} \right| \] First compute the dot product: \[ \vec{A} \cdot \vec{D} = (5\hat{i}) \cdot (2\hat{i} + 3\hat{j} - 6\hat{k}) = 10 \] Now compute the magnitude: \[ |\vec{D}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] Now compute the distance: \[ \text{Distance} = \left| \frac{10}{7} \right| = \frac{10}{7} \] Step 5: Scale to full displacement along direction
The vector \(\vec{A}\) was just from origin to one point on the plane. But we want the full perpendicular reach from origin to the plane, i.e., the scalar multiple \(\lambda\) of direction vector that lands on the plane. Instead of using projection on an arbitrary vector like \(5\hat{i}\), we scale the direction vector to reach the plane. In original method we got \(\lambda = -5\), so: \[ \text{Required distance} = |\lambda| \cdot |\vec{D}| = 5 \cdot 7 = 35 \] Final Answer: 35 units