Question

The distance of the point ( (5, 3, -1) ) from the plane passing through points ( (2, 1, 0), (3, -2, 4) ) and ( (1, -3, 3) ) is

• A. (\frac{2}{\sqrt{3}}) units
• B. (\frac{4}{\sqrt{3}}) units
• C. (\sqrt{3}) units
• D. [suspicious link removed] units

Hint:

Distance of $(x_1, y_1, z_1)$ from $Ax + By + Cz + D = 0$ is $\frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
First, find the equation of the plane passing through the three given points, then calculate the distance from the point to this plane.
Step 3: Detailed Explanation:
1. Points: \( A(2, 1, 0), B(3, -2, 4), C(1, -3, 3) \).
Normal vector \( \vec{n} = \vec{AB} \times \vec{AC} \):
\( \vec{AB} = (3-2, -2-1, 4-0) = (1, -3, 4) \)
\( \vec{AC} = (1-2, -3-1, 3-0) = (-1, -4, 3) \)
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -3 & 4
-1 & -4 & 3 \end{vmatrix} = \hat{i}(-9+16) - \hat{j}(3+4) + \hat{k}(-4-3) = 7\hat{i} - 7\hat{j} - 7\hat{k} \]
Direction ratios of normal: \( (1, -1, -1) \).
2. Plane equation: \( 1(x-2) - 1(y-1) - 1(z-0) = 0 \implies x - y - z - 1 = 0 \).
3. Distance from \( (5, 3, -1) \):
\[ d = \frac{|5 - 3 - (-1) - 1|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|5 - 3 + 1 - 1|}{\sqrt{3}} = \frac{2}{\sqrt{3}} \text{ units} \]
Step 4: Final Answer:
The distance is \( \frac{2}{\sqrt{3}} \) units.