3D Geometry
Distance of a Point from a Plane
Grade 12
Question:
<p>The distance of a point <math>\(2, 5, -3\)</math> from the plane <math>\(6x - 3y + 2z = 4\)</math> is</p>
<p>(a) <math>\(\frac{7}{13}\)</math></p>
<p>(b) <math>\(\frac{4}{13}\)</math></p>
<p>(c) <math>\(\frac{13}{7}\)</math></p>
<p>(d) <math>\(\frac{15}{7}\)</math></p>
Step-by-Step Solution
Key Concept: Apply the perpendicular distance formula from a point to a plane: substitute coordinates into the numerator and use the magnitude of the normal vector in the denominator.
Solution: Using the distance formula from a point <math>$(x_1, y_1, z_1)$</math> to the plane <math>$ax + by + cz + d = 0$</math>: Rewrite the plane as <math>$6x - 3y + 2z - 4 = 0$</math> Distance <math>$= \frac{|6(2) - 3(5) + 2(-3) - 4|}{\sqrt{6^2 + (-3)^2 + 2^2}} = \frac{|12 - 15 - 6 - 4|}{\sqrt{36 + 9 + 4}} = \frac{|-13|}{\sqrt{49}} = \frac{13}{7}$</math> ∴ Answer is (c) <math>$\frac{13}{7}$</math>
Correct Answer: c