Find the length and foot of the perpendicular segment from P(1,2,−3).

Open in App

Solution

$W e h a v e, L = \frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{Z}{- 1} = r, w h e r e P (1, 2, - 3) Q = f o o t o f t h e p e r p e n d i c u l a r (⊥) s o, Q = (2 r - 1, - 2 r + 3, - r) D r^{'} s o f P Q = (2 r - 2, - 2 r + 1, - r + 3) A n d, P Q ⊥ L t h e n, a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 2 (2 r - 2) - 2 (- 2 r + 1) - 1 (- r + 3) = 0 4 r - 4 - 4 r - 2 + r - 3 = 0 9 r - 9 = 0 ∴ r = 1 Q = (1, 1, - 1) P Q = \sqrt{{(1 - 1)}^{2} + {(1 - 2)}^{2} + {(- 1 + 3)}^{2}} = \sqrt{1 + 4} = \sqrt{5} S o, t h a t t h e t h e l e n g t h \sqrt{5} .$

Suggest Corrections

Similar questions

P (1, 2, - 3)

\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1}

P

(3, 4, 5)

(0, 2, 3)

\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{x + 4}{3}

\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}

Join BYJU'S Learning Program