Let the plane containing the line of intersection of the planes P1: x + ( + 4)y + z = 1 and P2: 2x + y + z = 2 pass thro

Question

Let the plane containing the line of intersection of the planes $P_1: x + (\lambda + 4)y + z = 1$ and $P_2: 2x + y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$ . Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane $P_2$ is:

A
$5\sqrt{6}$
B
$4\sqrt{6}$
C
$2\sqrt{6}$
D
$3\sqrt{6}$

Accepted Answer

Given: The required plane contains the line of intersection of $P_1: x + (\lambda + 4)y + z - 1 = 0$ and $P_2: 2x + y + z - 2 = 0$ , and it passes through $(0,1,0)$ and $(1,0,1)$ .

Find: The distance of the point $(2\lambda,\lambda,-\lambda)$ from the plane $P_2$ .

A plane through the line of intersection of two planes is

P = P_1 + kP_2 = 0

So,

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

which simplifies to

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

Since it passes through $(0,1,0)$ ,

\lambda + 4 + k - (1+2k) = 0

\lambda + 3 - k = 0

Also, since it passes through $(1,0,1)$ ,

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

1 + k = 0

Hence,

k = -1

Substituting $k = -1$ in $\lambda + 3 - k = 0$ ,

\lambda + 3 - (-1) = 0

\lambda + 4 = 0

\lambda = -4

Therefore the point becomes

(2\lambda,\lambda,-\lambda) = (-8,-4,4)

Now use the distance formula from the plane $P_2: 2x + y + z - 2 = 0$ :

d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2+b^2+c^2}}

So,

d = \frac{|2(-8) + (-4) + 4 - 2|}{\sqrt{2^2+1^2+1^2}}

d = \frac{|-18|}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3\sqrt{6}

Therefore, the distance is $3\sqrt{6}$ . The correct option is D.

The solution labels one place as option C, but the worked value is $3\sqrt{6}$ , which matches option D.

The key observation is that every plane through the line of intersection of $P_1$ and $P_2$ can be written as

P_1 + kP_2 = 0

Then the two given points are imposed to determine $k$ and $\lambda$ .

Using $(1,0,1)$ first gives a simpler equation:

(1+2k)(1) + (\lambda+4+k)(0) + (1+k)(1) - (1+2k) = 0

1+k=0

So,

k=-1

Then substitute into the condition from $(0,1,0)$ :

\lambda + 3 - k = 0

\lambda + 4 = 0

\lambda = -4

Now the point is

(-8,-4,4)

and distance from $2x+y+z-2=0$ is

\frac{|2(-8)+(-4)+4-2|}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3\sqrt{6}

Hence the correct option is D.

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