Question

Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is

• A. 6
• B. 4
• C. 5
• D. 7

Answer 1

The Correct Option is C

To find the distance of the plane \( P \) from the point \( (3, -2, 2) \), we must first determine the equation of the plane.

The plane \( P \) passes through the point \( (1, -1, -5) \) and is perpendicular to the line joining points \( (4, 1, -3) \) and \( (2, 4, 3) \).

First, we find the direction vector of the line joining the two points:

• The direction vector \( \mathbf{d} \) is given by \( \langle 2 - 4, 4 - 1, 3 - (-3) \rangle = \langle -2, 3, 6 \rangle \).

The normal vector to the plane is parallel to the direction vector of the line. Therefore, the normal vector \( \mathbf{n} \) to the plane is also \( \langle -2, 3, 6 \rangle \).

The equation of a plane with normal vector \( \mathbf{n} = \langle a, b, c \rangle \) and passing through point \( (x_0, y_0, z_0) \) is:

\(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\)

Substituting \( a = -2 \), \( b = 3 \), \( c = 6 \), and the point \( (1, -1, -5) \):

\(-2(x - 1) + 3(y + 1) + 6(z + 5) = 0\)

Expanding and simplifying gives:

\(-2x + 2 + 3y + 3 + 6z + 30 = 0\)

\(-2x + 3y + 6z + 35 = 0\)

The equation of the plane is:

\(2x - 3y - 6z = 35\)

Now, we find the distance of the point \( (3, -2, 2) \) from this plane using the formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( ax + by + cz + d = 0 \):

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

Substituting \( x_1 = 3 \), \( y_1 = -2 \), \( z_1 = 2 \), \( a = 2 \), \( b = -3 \), \( c = -6 \), and \( d = -35 \):

\(D = \frac{|2(3) - 3(-2) - 6(2) - 35|}{\sqrt{2^2 + (-3)^2 + (-6)^2}}\)

\(= \frac{|6 + 6 - 12 - 35|}{\sqrt{4 + 9 + 36}}\)

\(= \frac{|-35|}{\sqrt{49}}\)

\(= \frac{35}{7}\)

\(= 5\)

Therefore, the distance of the point \( (3, -2, 2) \) from the plane is 5.

The correct answer is 5.