Vector Algebra
Plane and point positions
Grade None
Question:
<p>The points \(\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}\) are equidistant from the plane \(\mathbf{r} \cdot (5\mathbf{i} + 2\mathbf{j} - 7\mathbf{k}) + 9 = 0\), then they are</p>
<p>(a) on the same sides of the plane</p>
<p>(b) parallel of the plane</p>
<p>(c) on the opposite sides of the plane</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Two points are equidistant from a plane if they lie on opposite sides (distances are equal but signs opposite) or same side (same sign). We check the signs of the plane equation evaluations at both points to determine their relative positions.
Step 1: Identify the plane equation and points. Plane: r · (5 i + 2 j - 7 k ) + 9 = 0, or 5x + 2y - 7z + 9 = 0 Point P_1 = i - j + 3 k = (1, -1, 3) Point P_2 = 3 i + 3 j + 3 k = (3, 3, 3) Step 2: Evaluate the plane equation at both points. At P_1: f(P_1) = 5(1) + 2(-1) - 7(3) + 9 = 5 - 2 - 21 + 9 = -9 At P_2: f(P_2) = 5(3) + 2(3) - 7(3) + 9 = 15 + 6 - 21 + 9 = 9 Step 3: Calculate perpendicular distances. Normal vector magnitude: | n | = √(25 + 4 + 49) = √78 Distance from P_1: d_1 = |−9|/√78 = 9/√78 Distance from P_2: d_2 = |9|/√78 = 9/√78 Therefore: d_1 = d_2 ✓ (equidistant condition satisfied) Step 4: Determine relative position using signs. f(P_1) = -9 < 0 (P_1 is on one side of the plane) f(P_2) = 9 > 0 (P_2 is on the opposite side of the plane) Since f(P_1) and f(P_2) have opposite signs, the points lie on opposite sides of the plane. ∴ Answer: The points are on opposite sides of the plane, so the answer is **(c)**. However, the problem statement indicates the correct answer is Unknown, suggesting there may be missing or ambiguous information in the original question formulation.
Correct Answer: Unknown