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 given line can be written in parametric form as: \[ x = 9 + 2t, \quad y = 13 + 3t, \quad z = 17 + 6t \] where \( t \) is the parameter. Let the point \( P(7, 10, 11) \) be the point from which we want to find the distance. The direction ratios of the line are \( 2, 3, 6 \), and the coordinates of the point on the line are \( (9, 13, 17) \). The distance \( D \) of the point \( P(x_1, y_1, z_1) \) from the line can be calculated using the formula: \[ D = \frac{| \vec{AP} \times \vec{d} |}{|\vec{d}|} \] where \( \vec{AP} = (x_1 - x_2, y_1 - y_2, z_1 - z_2) \) is the vector from a point on the line to the point \( P \), and \( \vec{d} = (2, 3, 6) \) is the direction vector of the line. Substitute the values: \[ \vec{AP} = (7 - 9, 10 - 13, 11 - 17) = (-2, -3, -6) \] \[ |\vec{d}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] Now, calculate the cross product \( \vec{AP} \times \vec{d} \): \[ \vec{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & -6 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}((-3)(6) - (-6)(3)) - \hat{j}((-2)(6) - (-6)(2)) + \hat{k}((-2)(3) - (-3)(2)) = \hat{i}(-18 + 18) - \hat{j}(-12 + 12) + \hat{k}(-6 + 6) = 0\hat{i} - 0\hat{j} + 0\hat{k} \] Thus, the magnitude of the cross product is: \[ |\vec{AP} \times \vec{d}| = \sqrt{0^2 + 0^2 + 0^2} = \sqrt{0} = 0 \] Finally, the distance is: \[ D = \frac{0}{7} = 0 \] Thus, the required distance is \( 0 \).