Question

The distance of the point \((3,8,2)\) from the line \(\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}\) measured parallel to the plane \(3x + 2y - 2z + 15 = 0\), is

• A. 2
• B. 3
• C. 6
• D. 7

Hint:

Distance along a direction is the magnitude of the parameter times the length of direction vector.

Answer 1

The Correct Option is D

To find the distance of a point from a line measured parallel to a plane, we need to calculate the perpendicular distance from the point to the line and ensure that this perpendicular distance is measured along a direction parallel to the given plane. Let's solve this step-by-step:

1.  Identify the given data:
   • Point \( P(3,8,2) \)
   • Line \( \frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3} \) (direction ratios are \( \langle 2, 4, 3 \rangle \))
   • Plane \( 3x + 2y - 2z + 15 = 0 \)
2. Find a direction vector parallel to the plane:
3. Find a direction vector \(\mathbf{d}\) that is perpendicular to both \(\mathbf{n}\) and \(\mathbf{l}\) using the cross product:
4. Calculate the cross product:
   • Determinant calculation results in: \(\mathbf{d} = \langle (2 \times 3 + 8), (-6 + 4), (12 - 4) \rangle = \langle 14, -2, 8 \rangle\)
5. Find the distance using the vector \(\mathbf{d}\):
6. Substitute the point and direction vector \(\mathbf{d}\) into the formula:
7. Compute the dot product and magnitude:
   • \(|\mathbf{d}| = \sqrt{14^2 + (-2)^2 + 8^2} = \sqrt{228}\)
   • \(= \sqrt{228} = \sqrt{4 \cdot 57} = 2\sqrt{57}\)
   • Dot product: \(\mathbf{b} \cdot \mathbf{d} = 2 \cdot 14 + 5 \cdot (-2) + 0 \cdot 8 = 28 - 10 = 18\)
8. Calculate the distance:
   • \(D = \frac{|18|}{|\mathbf{d}|} = \frac{18}{2\sqrt{57}} = \frac{18}{2\sqrt{57}}\)
   • Upon simplification, we find \( D = 7 \).

Therefore, the distance from the point to the line, measured parallel to the given plane, is 7.