Distance of the point (2, 4, −1) from the line \(\displaystyle\frac{10 + 2x}{2} = \frac{y + 3}{4} = \frac{6 - z}{9}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7 units
In standard form, with the coefficients of x, y and z equal to 1:
\(\displaystyle \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} = \lambda\)
A general point of the line is Q(λ − 5, 4λ − 3, 6 − 9λ). For P(2, 4, −1):
\(\vec{PQ} = (\lambda - 7)\hat{\imath} + (4\lambda - 7)\hat{\jmath} + (7 - 9\lambda)\hat{k}\)
Q is the foot of the perpendicular when \(\vec{PQ}\) is perpendicular to the direction \(\hat{\imath} + 4\hat{\jmath} - 9\hat{k}\):
\((\lambda - 7) + 4(4\lambda - 7) - 9(7 - 9\lambda) = 0\)
\(98\lambda - 98 = 0,\quad \lambda = 1\)
So the foot is Q(−4, 1, −3), and the distance is
\(PQ = \sqrt{(2 + 4)^2 + (4 - 1)^2 + (-1 + 3)^2}\)
\(= \sqrt{36 + 9 + 4} = 7\) units
NCERT Exemplar:
Three Dimensional Geometry
