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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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:

142

No worries! We‘ve got your back. Try BYJU‘S free classes today!

68

No worries! We‘ve got your back. Try BYJU‘S free classes today!

136

No worries! We‘ve got your back. Try BYJU‘S free classes today!

102

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $102$
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
$21 (α + β + γ) = 21 (\frac{34}{7}) = 102$

Suggest Corrections

Similar questions

Q. Let

L

be the line of intersection of planes

\to r \cdot (^i -^j + 2^k) = 2

and

\to r \cdot (2^i +^j -^k) = 2

. If

P (α, β, γ)

is the foot of perpendicular on

L

from the point

(1, 2, 0)

, then the value of

35 (α + β + γ)

is equal to

Q. If two lines

\frac{x - α}{l} = \frac{y - β}{m} = \frac{z - γ}{n}

and

\frac{x - α^{'}}{l^{'}} = \frac{y - β^{'}}{m^{'}} = \frac{z - γ^{'}}{n^{'}}

lie in the same plane then the equation of the plane is -

Q. Let a plane

P

contain two lines

\to r =^i + λ (^i +^j), λ \in R

and

\to r = -^j + μ (^j -^k), μ \in R .

Q (α, β, γ)

is the foot of the perpendicular drawn from the point

M (1, 0, 1)

P

, then

3 (α + β + γ)

equals

Q. Determine the equation of the plane on which the co - ordinates of the foot of perpendicular drawn from origin O is the point

P (α, β, γ) .

Q. Let the equation of the plane, that passes through the point

(1, 4, - 3)

and contains the line of intersection of the planes

3 x - 2 y + 4 z - 7 = 0

and

x + 5 y - 2 z + 9 = 0,

α x + β y + γ z + 3 = 0,

then

α + β + γ

is equal to

Join BYJU'S Learning Program

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 D 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: