Question

The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is

• A. 4 units
• B. 5 units
• C. 6 units
• D. 7 units

Hint:

To find the perpendicular distance from a point to a line, first parameterize the line. Then, form a vector from the given point to a general point on the line. The dot product of this vector and the line's direction vector must be zero for perpendicularity. Solve for the parameter, find the specific point on the line, and then calculate the distance.

Answer 1

The Correct Option is D

Step 1: Parameterize the given line. Let the equation of the line be $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$. Any point $P$ on this line can be represented by its coordinates in terms of $\lambda$: \[ P = (3\lambda + 6, 2\lambda + 7, -2\lambda + 7) \]
Step 2: Define the given point. Let the given point be $A = (1, 2, 3)$.
Step 3: Find the direction ratios of the line segment $AP$. The direction ratios of $AP$ are given by the differences in coordinates: \[ (x_P - x_A, y_P - y_A, z_P - z_A) = ((3\lambda + 6) - 1, (2\lambda + 7) - 2, (-2\lambda + 7) - 3) \] \[ = (3\lambda + 5, 2\lambda + 5, -2\lambda + 4) \]
Step 4: Use the condition for perpendicularity. The line segment $AP$ is perpendicular to the given line. The direction ratios of the given line are $(3, 2, -2)$. For two lines to be perpendicular, the dot product of their direction ratios must be zero. \[ (3\lambda + 5)(3) + (2\lambda + 5)(2) + (-2\lambda + 4)(-2) = 0 \] \[ 9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \] \[ (9\lambda + 4\lambda + 4\lambda) + (15 + 10 - 8) = 0 \] \[ 17\lambda + 17 = 0 \] \[ 17\lambda = -17 \] \[ \lambda = -1 \]
Step 5: Find the coordinates of point $P$. Substitute $\lambda = -1$ back into the coordinates of point $P$: \[ P = (3(-1) + 6, 2(-1) + 7, -2(-1) + 7) \] \[ P = (-3 + 6, -2 + 7, 2 + 7) \] \[ P = (3, 5, 9) \]
Step 6: Calculate the distance $AP$. The length of the perpendicular is the distance between point $A(1, 2, 3)$ and point $P(3, 5, 9)$. Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$: \[ AP = \sqrt{(3-1)^2 + (5-2)^2 + (9-3)^2} \] \[ AP = \sqrt{(2)^2 + (3)^2 + (6)^2} \] \[ AP = \sqrt{4 + 9 + 36} \] \[ AP = \sqrt{49} \] \[ AP = 7 \text{ units} \]