Question

The distance of the point \( (7, 10, 11) \) from the line \( \frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6} \) along the line is:

• A. \( \frac{1}{\sqrt{14}} \)
• B. \( \frac{2}{\sqrt{14}} \)
• C. \( \frac{3}{\sqrt{14}} \)
• D. \( \frac{4}{\sqrt{14}} \)

Hint:

Use the formula for the distance from a point to a line in three-dimensional space involving the cross product and the direction ratios of the line.

Answer 1

The Correct Option is B

The parametric representation of the given line is: \( x = 9 + 2t, \quad y = 13 + 3t, \quad z = 17 + 6t \), where \( t \) is the parameter. We aim to find the distance from the point \( P(7, 10, 11) \). The line has direction ratios \( 2, 3, 6 \), and a point on the line is \( (9, 13, 17) \). The distance \( D \) from a point \( P(x_1, y_1, z_1) \) to the line is given by: \( D = \frac{| \vec{AP} \times \vec{d} |}{|\vec{d}|} \). Here, \( \vec{AP} = (x_1 - x_2, y_1 - y_2, z_1 - z_2) \) is the vector from a point on the line to \( P \), and \( \vec{d} = (2, 3, 6) \) is the direction vector of the line. Substituting the values: \( \vec{AP} = (7 - 9, 10 - 13, 11 - 17) = (-2, -3, -6) \). The magnitude of the direction vector is: \( |\vec{d}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \). The cross product \( \vec{AP} \times \vec{d} \) is calculated as: \( \vec{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & -6 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}((-3)(6) - (-6)(3)) - \hat{j}((-2)(6) - (-6)(2)) + \hat{k}((-2)(3) - (-3)(2)) = \hat{i}(-18 + 18) - \hat{j}(-12 + 12) + \hat{k}(-6 + 6) = 0\hat{i} - 0\hat{j} + 0\hat{k} \). The magnitude of this cross product is: \( |\vec{AP} \times \vec{d}| = \sqrt{0^2 + 0^2 + 0^2} = \sqrt{0} = 0 \). Therefore, the distance is: \( D = \frac{0}{7} = 0 \). The required distance is \( 0 \).