3D Geometry
Planes and Distance Formula
Grade 12
Question:
<p>The distance of the point <span class="math">(1, 3, -7)</span> from the plane passing through the point <span class="math">(1, -1, -1)</span> having normal perpendicular to both the lines <span class="math">\frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-4}{3}</span> and <span class="math">\frac{x-2}{2} = \frac{y+1}{-1} = \frac{z+7}{-1}</span> is</p><p>(JEE Main 2017)</p>
<p>(a) <span class="math">\frac{20}{\sqrt{74}}</span> units</p>
<p>(b) <span class="math">\frac{10}{\sqrt{83}}</span> units</p>
<p>(c) <span class="math">\frac{5}{\sqrt{83}}</span> units</p>
<p>(d) <span class="math">\frac{10}{\sqrt{74}}</span> units</p>
Step-by-Step Solution
Key Concept: The distance from a point to a plane is found using the formula involving the dot product of the point vector with the normal and the magnitude of the normal. The normal to the plane is perpendicular to both given lines, found via cross product.
Solution: Given equations of lines: \frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-4}{3} and \frac{x-2}{2} = \frac{y+1}{-1} = \frac{z+7}{-1} Direction vectors of the lines are: \vec{n}_1 = \hat{i} - 2\hat{j} + 3\hat{k} \vec{n}_2 = 2\hat{i} - \hat{j} - \hat{k} The normal to the plane perpendicular to both lines is: \vec{n} = \vec{n}_1 \times \vec{n}_2 Using the cross product formula and the point-to-plane distance formula: d = \frac{|\vec{AP} \cdot \vec{n}|}{|\vec{n}|} where A = (1, -1, -1) and P = (1, 3, -7). Computing the cross product and distance yields \frac{10}{\sqrt{83}} units. ∴ Answer is (b)
Correct Answer: B