Let the plane containing the line of intersection of the planes P 1: x+(λ+4) y+z=1 and P 2: 2 x+ y+z=2 pass through the points (0,1,0) and (1,0,1). Then the distance of the point (2 λ, λ,-λ) from the plane P 2 is

Question

Let the plane containing the line of intersection of the planes P 1: x+(λ+4) y+z=1 and P 2: 2 x+ y+z=2 pass through the points (0,1,0) and (1,0,1). Then the distance of the point (2 λ, λ,-λ) from the plane P 2 is (A) 5 √6 (B) 2 √6 (C) 3 √6 (D) 4 √6

Tardigrade Admin · Accepted Answer

Equation of plane passing through point of intersection of

P 1

and

P 2

P = P 1 + k P 2

(x + (λ + 4) y + z - 1) + k (2 x + y + z - 2) = 0

Passing through

(0, 1, 0)

and

(1, 0, 1)

(λ + 4 - 1) + k (1 - 2) = 0

(λ + 3) - k = 0.....

(1)
Also passing

(1, 0, 1)

(1 + 1 - 1) + k (2 + 1 - 2) = 0

1 + k = 0

k = - 1

put in (1)

λ + 3 + 1 = 0

λ = - 4

Then point

(2 λ, λ, - λ)

d = ∣ ∣ \frac{- 16 - 4 , - 4 , 4 )}{6} ∣ ∣

d = \frac{18}{6} \times \frac{6}{6} = 36

Q. Let the plane containing the line of intersection of the planes $P 1 : x + (λ + 4) y + z = 1$ and $P 2 : 2 x +$ $y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$ . Then the distance of the point $(2 λ, λ, - λ)$ from the plane $P 2$ is

$56$

$26$

$36$

$46$

Q. Let the plane containing the line of intersection of the planes P1:x+(λ+4)y+z=1 and P2:2x+ y+z=2 pass through the points (0,1,0) and (1,0,1). Then the distance of the point (2λ,λ,−λ) from the plane P2 is

56​

26​

36​

46​

Q. Let the plane containing the line of intersection of the planes $P 1 : x + (λ + 4) y + z = 1$ and $P 2 : 2 x +$ $y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$ . Then the distance of the point $(2 λ, λ, - λ)$ from the plane $P 2$ is

$56$

$26$

$36$

$46$