If the foot of the perpendicular from the point A(-1,4,3) on the plane P: 2 x+m y+n z=4, is (-2, 7/2, 3/2), then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3,-1,-4, is equal to :

Solution

Maths sir · 5 steps

Read the normal off the foot of the perpendicular, then slide along the given line until you hit the plane.

1. Find the normal. The segment from A(-1,4,3) to its foot F(-2,7/2,3/2) is perpendicular to the plane. AF⃗ = (-1,-1/2,-3/2), which is parallel to (2,1,3).
2. Fix the plane. The normal (2,m,n) must be parallel to (2,1,3), so m = 1 and n = 3. Check with F: 2(-2) + 7/2 + 3(3/2) = 4. So P is 2x + y + 3z = 4.
3. Line through A. A general point on the line through A with direction ratios 3,-1,-4 is (-1+3t, 4-t, 3-4t). Put it into the plane to find where it meets P.

   2(-1+3t) + (4-t) + 3(3-4t) = 4 ⇒ 11 - 7t = 4 ⇒ t = 1

4. Measure the distance. The line meets P at B(2,3,-1). Then AB = |t|√(3² + 1² + 4²) = 1 × √26.
5. So. The distance of A from P measured parallel to the given line is √26, option (b).

The idea to keep. Distance measured parallel to a line means: go from the point along that line to the plane and measure the segment. The parameter t is not the distance unless the direction vector is a unit vector.

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