<p>The coordinates of the foot of the perpendicular drawn from the origin to the plane <i>2x + 3y + 4z − 12 = 0</i> is</p>
<p>(a) <i>(24/29, 36/29, 48/29)</i></p>
<p>(b) <i>(48/29, 36/29, 24/29)</i></p>
<p>(c) <i>(24/29, 36/29, 48/29)</i></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The foot of perpendicular from a point to a plane lies on the line through that point parallel to the normal vector of the plane. The normal vector (2, 3, 4) from the plane equation gives us the direction, and we find where this line intersects the plane.
Step 1: Identify the normal vector to the plane 2x + 3y + 4z − 12 = 0. The normal vector is n = (2, 3, 4). Step 2: The foot of perpendicular from origin O(0, 0, 0) lies on the line passing through O in the direction of the normal vector. Parametric equation of this line: (x, y, z) = (0, 0, 0) + t(2, 3, 4) = (2t, 3t, 4t) Step 3: This line intersects the plane 2x + 3y + 4z − 12 = 0. Substitute the parametric equations into the plane equation: 2(2t) + 3(3t) + 4(4t) − 12 = 0 Step 4: Simplify: 4t + 9t + 16t − 12 = 0 → 29t = 12 → t = 12/29 Step 5: The foot of perpendicular is at (2t, 3t, 4t) = (2·12/29, 3·12/29, 4·12/29) = (24/29, 36/29, 48/29) Step 6: Verify: 2(24/29) + 3(36/29) + 4(48/29) = (48 + 108 + 192)/29 = 348/29 = 12 ✓ ∴ Answer: A
Correct Answer: A