Question

The distance of a point \( (2,3,-5) \) from the plane \( \vec{r} \cdot (4i - 3j + 2k) = 4 \) is:

• A. \( \frac{11}{2} \)
• B. \( \frac{11}{\sqrt{29}} \)
• C. \( \frac{15}{\sqrt{29}} \)
• D. \( \frac{11}{\sqrt{38}} \)

Hint:

Use the point-to-plane distance formula directly to avoid unnecessary calculations.

Answer 1

The Correct Option is C

The distance \( d \) of a point \( (x_1, y_1, z_1) \) from a plane \( ax + by + cz = d \) is given by the formula:
\( d = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}} \)
For the point \( (2, 3, -5) \) and the plane \(\vec{r} \cdot (4i - 3j + 2k) = 4 \), the plane equation is:
\( 4x - 3y + 2z = 4 \).
Substitute the point into the distance formula:
\( d = \frac{|4 \times 2 - 3 \times 3 + 2 \times (-5) - 4|}{\sqrt{4^2 + (-3)^2 + 2^2}} \)
\( d = \frac{|8 - 9 - 10 - 4|}{\sqrt{16 + 9 + 4}} \)
Simplifying the numerator:
\( d = \frac{|-15|}{\sqrt{29}} \)
\( d = \frac{15}{\sqrt{29}} \)

Answer 2

Step 1: Use the distance formula for a point to a plane The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by: \[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Step 2: Substitute values Given the plane equation \( 4x - 3y + 2z = 4 \), the coefficients are: - \( A = 4, B = -3, C = 2, D = -4 \) - Point \( (2,3,-5) \) \[ d = \frac{|4(2) -3(3) + 2(-5) - 4|}{\sqrt{4^2 + (-3)^2 + 2^2}} \] \[ = \frac{|8 - 9 - 10 - 4|}{\sqrt{16 + 9 + 4}} \] \[ = \frac{15}{\sqrt{29}} \] Thus, the correct answer is option (3).