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

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: $PQ = 13.$