3D Geometry
Equation of Plane
Grade 12
Question:
<p>Consider the plane through \((2, 3, -1)\) and at right angles to the vector \(3\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\) from the origin is</p>
<p>(a) The equation of the plane through the given point is \(3x - 4y + 7z + 13 = 0\)</p>
<p>(b) perpendicular distance of plane from origin \(= \frac{13}{\sqrt{74}}\)</p>
<p>(c) perpendicular distance of plane from origin</p>
<p>(d) Other option</p>
Step-by-Step Solution
Key Concept: Use the normal vector and point to form the plane equation, then calculate distance from origin.
Step 1: The plane perpendicular to vector \(\mathbf{n} = 3\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\) and passing through \((2, 3, -1)\) has equation \(3(x-2) - 4(y-3) + 7(z+1) = 0\). Step 2: Expanding: \(3x - 6 - 4y + 12 + 7z + 7 = 0\), which gives \(3x - 4y + 7z + 13 = 0\). Step 3: Distance from origin \(= \frac{|13|}{\sqrt{9+16+49}} = \frac{13}{\sqrt{74}}\).
Correct Answer: A