If foot of the perpendicular of P 2-3-1 on the line x -1-2- y -3-3- z -2-1 is Q a - b - c then find the value of 14 a - b - c

Given line is x+1/2 = y 3/3=z+2/ 1=r....(1) Point P ≡ (2, 3,1) Co ordinates of foot of the perpendicular on line (1) may be taken as Q ≡ (2r 1 ...

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Byju's Answer

Standard XII

Mathematics

Foot of Perpendicular of a Point on a Line

If foot of th...

Question

If foot of the perpendicular of P(2,-3,1) on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{- 1}$ is Q(a,b,c) then find the value of 14(a+b+c)

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Solution

$G i v e n l i n e i s \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{- 1} = r . . . . (1) P o i n t P \equiv (2, - 3, 1) C o - o r d i n a t e s o f f o o t o f t h e p e r p e n d i c u l a r o n l i n e (1) m a y b e t a k e n a s Q \equiv (2 r - 1, 3 r + 3, - r - 2) W e g e t d i r e c t i o n r a t i o s o f P Q = 2 r - 3, 3 r + 6, - r - 3 D i r e c t i o n r a t i o s o f l i n e s e g m e n t a r e 2, 3, - 1 (f r o m (1) S i n c e P Q p e r p e n d i c u l a r t o A B ∴ 2 (2 r - 3) + 3 (3 r + 6) - 1 (- r - 3) = 0 o r, 14 r + 15 = 0 ∴ r = \frac{- 15}{14} ∴ Q \equiv (2 r - 1, 3 r + 3, - r - 2) = (\frac{- 22}{7}, \frac{- 3}{14}, \frac{- 13}{14}) = (a, b, c) (g i v e n) \Rightarrow - 14 (a + b + c) = - 14 (\frac{- 22}{7} \frac{- 13}{14}, \frac{- 13}{14}) = 44 + 3 + 13 = 60$

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Similar questions

If foot of the perpendicular of P(2,-3,1) on the line
$\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{- 1}$ is
$Q (a, b, c)$ then find the value of $- 14 (a + b + c)$

Q. Find the length of perpendicular from

P (2, - 3, 1)

to the line

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

Q. The foot of the perpendicular of P(2, -3, 1) on the line

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

is Q. If the DRs of the line segment joining P and Q is l, m, n, then .

Q. Find the length of the perpendicular from

(2, - 3, 1)

to the line

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

Let the foot of the perpendicular of $P (2, - 3, 1)$ on the line
$\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 2}{- 1}$ be $Q .$
If direction ratios of the line segment joining P and Q be $l, m, n$ , then which of the following relations are correct ?

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If foot of the perpendicular of P(2,-3,1) on the line x+12=y−33=z+2−1 is Q(a,b,c) then find the value of 14(a+b+c) ___

If foot of the perpendicular of P(2,-3,1) on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{- 1}$ is Q(a,b,c) then find the value of 14(a+b+c)

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