Let L be a line obtained from the intersection of two planes x-2y-z-6 and y-2z-4- If point P- is the foot of perpendicular from 3- 2- 1 on L- then the value of 21- equals-

Equation of the line x+2y+z 6=0=y+2z=4 or x+23=y 4 2=z1=λ D.r.'s of PQ:3λ 5, 2λ 2, λ 1 D.r.'s of y lines are (3, 2,1) Since PQ ⊥ line 3(3λ 5) 2( 2λ+2)+1(λ 1)=0 ...

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Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

Let L be a li...

Question

Let $L$ be a line obtained from the intersection of two planes $x + 2 y + z = 6$ and $y + 2 z = 4$ . If point $P (α, β, γ)$ is the foot of perpendicular from $(3, 2, 1)$ on $L$ , then the value of $21 (α + β + γ)$ equals:

102

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Solution

The correct option is A $102$
Equation of the line $x + 2 y + z - 6 = 0 = y + 2 z = 4$
or $\frac{x + 2}{3} = \frac{y - 4}{- 2} = \frac{z}{1} = λ$

D.r.'s of $P Q : 3 λ - 5, - 2 λ - 2, λ - 1$
D.r.'s of $y$ lines are $(3, - 2, 1)$
Since $P Q ⊥$ line
$3 (3 λ - 5) - 2 (- 2 λ + 2) + 1 (λ - 1) = 0$
$\Rightarrow λ = \frac{10}{7}$
$∴ P \equiv (\frac{16}{7}, \frac{8}{7}, \frac{10}{7}) \equiv (α, β, γ)$
Hence
$21 (α + β + γ) = 21 (\frac{34}{7}) = 102$

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Let L be a line obtained from the intersection of two planes x+2y+z=6 and y+2z=4. If point P(α,β,γ) is the foot of perpendicular from (3,2,1) on L, then the value of 21(α+β+γ) equals:

The correct option is A 102Equation of the line x+2y+z−6=0=y+2z=4 or x+23=y−4−2=z1=λ D.r.'s of PQ:3λ−5,−2λ−2,λ−1 D.r.'s of y lines are (3,−2,1) Since PQ⊥ line 3(3λ−5)−2(−2λ+2)+1(λ−1)=0 ⇒λ=107 ∴P≡(167,87,107)≡(α,β,γ) Hence 21(α+β+γ)=21(347)=102

Let $L$ be a line obtained from the intersection of two planes $x + 2 y + z = 6$ and $y + 2 z = 4$ . If point $P (α, β, γ)$ is the foot of perpendicular from $(3, 2, 1)$ on $L$ , then the value of $21 (α + β + γ)$ equals: