Question

The distance of a point \( \vec{a} \) from the plane \( r \cdot m = q \) is given by \( \frac{| \vec{a} \cdot m - q |}{|m|} \). If the distance of the point \( i + 2j + 3k \) from the plane \( \vec{r} \cdot (2i + 6j - 9k) = -1 \) is \( p \) and the distance of the origin from this plane is \( q \), then \( p - q = \dots \)

• A. 6
• B. 5
• C. 2
• D. 1

Hint:

In problems involving the distance of points from planes, always start by using the formula for the distance and carefully compute dot products and magnitudes.

Answer 1

The Correct Option is D

We are given the following: - The equation of the plane: \( \vec{r} \cdot (2i + 6j - 9k) = -1 \) - The point: \( \vec{a} = i + 2j + 3k \) - The distance of the point from the plane is \( p \), and the distance of the origin from the plane is \( q \). Step 1: Use the formula for the distance of a point from a plane: \[ \text{Distance} = \frac{| \vec{a} \cdot \vec{m} - q |}{|\vec{m}|} \] where \( \vec{m} = 2i + 6j - 9k \). Step 2: Calculate the dot product \( \vec{a} \cdot \vec{m} \): \[ \vec{a} \cdot \vec{m} = (1)(2) + (2)(6) + (3)(-9) = 2 + 12 - 27 = -13 \] So, \( \vec{a} \cdot \vec{m} - q = -13 + 1 = -12 \). Step 3: Calculate the magnitude of \( \vec{m} \): \[ |\vec{m}| = \sqrt{2^2 + 6^2 + (-9)^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11 \] Step 4: The distance \( p \) is: \[ p = \frac{|-12|}{11} = \frac{12}{11} \] Step 5: The distance of the origin from the plane is \( q \), and we know that \( q = \frac{|1|}{11} = \frac{1}{11} \). Step 6: The required difference is: \[ p - q = \frac{12}{11} - \frac{1}{11} = \frac{11}{11} = 1 \] % Final Answer \[ \boxed{1} \]

Answer 2

Given:
We are to compute the value of \( p - q \), where:
- \( p \) is the distance from point \( \vec{a} = i + 2j + 3k \) to the plane \( \vec{r} \cdot (2i + 6j - 9k) = -1 \)
- \( q \) is the distance from the origin to the same plane

Formula for Distance from a Point to a Plane:
The distance \( D \) of a point \( \vec{a} \) from a plane \( \vec{r} \cdot \vec{m} = c \) is:
\[ D = \frac{|\vec{a} \cdot \vec{m} - c|}{|\vec{m}|} \]

Step 1: Find vector \( \vec{m} \)
The normal vector to the plane is \( \vec{m} = 2i + 6j - 9k \)
So, \( |\vec{m}| = \sqrt{2^2 + 6^2 + (-9)^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11 \)

Step 2: Find p — distance from point \( i + 2j + 3k \)
Let \( \vec{a} = i + 2j + 3k \)
Then, \( \vec{a} \cdot \vec{m} = (1)(2) + (2)(6) + (3)(-9) = 2 + 12 - 27 = -13 \)
Plane equation: \( \vec{r} \cdot \vec{m} = -1 \Rightarrow c = -1 \)
\[ p = \frac{| -13 - (-1) |}{11} = \frac{|-13 + 1|}{11} = \frac{12}{11} \]

Step 3: Find q — distance from the origin
Origin \( \vec{a} = 0 \), so \( \vec{a} \cdot \vec{m} = 0 \)
\[ q = \frac{|0 - (-1)|}{11} = \frac{1}{11} \]

Step 4: Compute p - q
\[ p - q = \frac{12}{11} - \frac{1}{11} = \frac{11}{11} = 1 \]

Final Answer:
\( p - q = \boxed{1} \)