The distance of the line (x-2)/2=(y-6)/3=(z-3)/4 from the point (1,4,0) along the line x/1=(y-2)/2=(z+3)/3 is :

Solution

Maths sir · 4 steps

Walk from the point along the given direction until you hit the first line, then measure that walk.

1. What is asked. This is not the perpendicular distance. Start at (1,4,0), move parallel to (1,2,3), and find where you meet the first line. Your point lies on the second line: 1/1 = (4-2)/2 = (0+3)/3 = 1.
2. General point. Any point on the path is P = (1+t, 4+2t, 3t). Put it into the first line's ratios and solve for t.

   ((1+t)-2)/2 = ((4+2t)-6)/3 = (3t-3)/4 ⇒ (t-1)/2 = 2(t-1)/3 = 3(t-1)/4

3. Find the meeting point. The three fractions 1/2, 2/3, 3/4 are different, so all three ratios are equal only when t - 1 = 0. So t = 1, and the meeting point is (2, 6, 3).
4. So. The distance is √((2-1)² + (6-4)² + (3-0)²) = √(1+4+9) = √14, which is option (d).

The idea to keep. "Distance along a line" means move along that direction until you hit the target, then measure. Parametrise the path, put it into the target's equations, solve for the parameter.

Answer from the NTA final answer key. Solution by PrepWiser.