Question

Let P be the point (10, –2, –1) and Q be the foot of the perpendicular drawn from the point R(1, 7, 6) on the line passing through the points (2, –5, 11) and (–6, 7, –5). Then the length of the line segment PQ is equal to ________.

Answer 1

Correct Answer: 13

The equation of the line passing through \( (2, -5, 11) \) and \( (-6, 7, -5) \) is:

\[ \frac{x + 6}{-8} = \frac{y - 7}{12} = \frac{z + 5}{-16} = \lambda. \]

Using the parametric form of the line:

\[ Q(\lambda) = (2\lambda - 6, 7 - 3\lambda, 4\lambda - 5). \]

The vector \( QR \) (from \( Q \) to \( R(1, 7, 6) \)) is given as:

\[ QR = (2\lambda - 7, -3\lambda, 4\lambda - 11). \]

Since \( QR \) is perpendicular to the line, the direction ratios \((-8, 12, -16)\) of the line satisfy:

\[ -8(2\lambda - 7) + 12(-3\lambda) + (-16)(4\lambda - 11) = 0. \]

Simplify:

\[ -16\lambda + 56 - 36\lambda - 64\lambda + 176 = 0. \]

Combine terms:

\[ 116\lambda = 232 \implies \lambda = 2. \]

Substitute \( \lambda = 2 \) in the parametric equation of the line:

\[ Q(2) = (2(2) - 6, 7 - 3(2), 4(2) - 5) = (-2, 1, 3). \]

The length of the segment \( PQ \) is:

\[ PQ = \sqrt{(10 - (-2))^2 + (-2 - 1)^2 + (-1 - 3)^2}. \]

Simplify:

\[ PQ = \sqrt{(12)^2 + (-3)^2 + (-4)^2} = \sqrt{144 + 9 + 16} = \sqrt{169}. \]

Thus: \[ PQ = 13. \]

Answer 2

The problem asks for the length of the line segment PQ. We are given the coordinates of point P as (10, –2, –1). The point Q is the foot of the perpendicular drawn from point R(1, 7, 6) to the line that passes through the points A(2, –5, 11) and B(–6, 7, –5).

Concept Used:

To solve this problem, we will use concepts from three-dimensional coordinate geometry:

1. Equation of a Line in 3D: The equation of a line passing through a point \( \vec{a} \) with a direction vector \( \vec{d} \) is given by \( \vec{r} = \vec{a} + \lambda \vec{d} \). The direction vector can be found by taking the difference between the position vectors of two points on the line.

2. Foot of the Perpendicular: Let Q be the foot of the perpendicular from a point R to a line. The vector connecting R and Q, \( \vec{RQ} \), will be perpendicular to the direction vector of the line. The condition for perpendicularity of two vectors is that their dot product is zero.

3. Distance Formula in 3D: The distance between two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) is given by:

\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Step-by-Step Solution:

Step 1: Find the direction vector and the equation of the line passing through A(2, –5, 11) and B(–6, 7, –5).

The direction vector \( \vec{d} \) of the line is given by the vector \( \vec{AB} \):

\[ \vec{d} = \vec{B} - \vec{A} = (-6 - 2)\hat{i} + (7 - (-5))\hat{j} + (-5 - 11)\hat{k} \] \[ \vec{d} = -8\hat{i} + 12\hat{j} - 16\hat{k} \]

We can simplify this direction vector by dividing by their greatest common divisor, -4, to get a simpler parallel vector \( \vec{d'} \):

\[ \vec{d'} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]

The equation of the line passing through point A(2, –5, 11) with this direction vector is:

\[ \frac{x - 2}{2} = \frac{y + 5}{-3} = \frac{z - 11}{4} = \lambda \]

Step 2: Determine the coordinates of a general point Q on this line.

From the line equation, any point Q on the line can be represented in terms of the parameter \( \lambda \):

\[ x = 2\lambda + 2 \] \[ y = -3\lambda - 5 \] \[ z = 4\lambda + 11 \]

So, the coordinates of Q are \( (2\lambda + 2, -3\lambda - 5, 4\lambda + 11) \).

Step 3: Find the direction vector of the line segment RQ.

The coordinates of R are (1, 7, 6). The direction vector \( \vec{RQ} \) is \( \vec{Q} - \vec{R} \):

\[ \vec{RQ} = (2\lambda + 2 - 1)\hat{i} + (-3\lambda - 5 - 7)\hat{j} + (4\lambda + 11 - 6)\hat{k} \] \[ \vec{RQ} = (2\lambda + 1)\hat{i} + (-3\lambda - 12)\hat{j} + (4\lambda + 5)\hat{k} \]

Step 4: Apply the condition of perpendicularity to find \( \lambda \).

Since Q is the foot of the perpendicular from R, the vector \( \vec{RQ} \) is perpendicular to the direction vector of the line, \( \vec{d'} \). Their dot product must be zero:

\[ \vec{RQ} \cdot \vec{d'} = 0 \] \[ (2\lambda + 1)(2) + (-3\lambda - 12)(-3) + (4\lambda + 5)(4) = 0 \] \[ 4\lambda + 2 + 9\lambda + 36 + 16\lambda + 20 = 0 \] \[ (4 + 9 + 16)\lambda + (2 + 36 + 20) = 0 \] \[ 29\lambda + 58 = 0 \implies 29\lambda = -58 \implies \lambda = -2 \]

Step 5: Find the coordinates of the point Q.

Substitute \( \lambda = -2 \) into the general coordinates of Q:

\[ x_Q = 2(-2) + 2 = -4 + 2 = -2 \] \[ y_Q = -3(-2) - 5 = 6 - 5 = 1 \] \[ z_Q = 4(-2) + 11 = -8 + 11 = 3 \]

Thus, the coordinates of Q are (–2, 1, 3).

Final Computation & Result:

Step 6: Calculate the length of the line segment PQ using the distance formula.

The coordinates are P(10, –2, –1) and Q(–2, 1, 3).

\[ PQ = \sqrt{(-2 - 10)^2 + (1 - (-2))^2 + (3 - (-1))^2} \] \[ PQ = \sqrt{(-12)^2 + (3)^2 + (4)^2} \] \[ PQ = \sqrt{144 + 9 + 16} \] \[ PQ = \sqrt{169} \] \[ PQ = 13 \]

The length of the line segment PQ is 13.