Question

The shortest distance between the point \( (2,3,4) \) and the line \( \dfrac{x-4}{-2}=\dfrac{y-4}{2}=\dfrac{z-6}{1} \) is

• A. \( 12 \)
• B. \( 9 \)
• C. \( 3 \)
• D. \( \sqrt{5} \)
• E. \( \sqrt{3} \)

Hint:

For the shortest distance from a point to a line in 3D, always use \( \dfrac{|\vec{AP}\times\vec{d}|}{|\vec{d}|} \), where \( \vec{d} \) is the direction vector of the line.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the shortest distance from a point to a line in 3D space. This is the length of the perpendicular from the point to the line.
Step 2: Key Formula or Approach:
Let the point be P and the line be L. Let A be a point on the line L and \(\vec{b}\) be the direction vector of the line. The shortest distance \(d\) from point P to line L is given by the formula:
\[ d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|} \] Step 3: Detailed Explanation:
From the given information:
The point is \(P(2, 3, 4)\).
The line L is \( \frac{x-4}{-2} = \frac{y-4}{2} = \frac{z-6}{1} \).
From the equation of the line, we can identify a point on the line and its direction vector.
A point on the line is \(A(4, 4, 6)\).
The direction vector is \(\vec{b} = -2\mathbf{i} + 2\mathbf{j} + 1\mathbf{k}\).
Now, we find the vector \(\vec{AP}\):
\[ \vec{AP} = \vec{P} - \vec{A} = (2-4)\mathbf{i} + (3-4)\mathbf{j} + (4-6)\mathbf{k} = -2\mathbf{i} - 1\mathbf{j} - 2\mathbf{k} \] Next, we calculate the cross product \(\vec{AP} \times \vec{b}\):
\[ \vec{AP} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
-2 & -1 & -2
-2 & 2 & 1 \end{vmatrix} \] \[ = \mathbf{i}((-1)(1) - (-2)(2)) - \mathbf{j}((-2)(1) - (-2)(-2)) + \mathbf{k}((-2)(2) - (-1)(-2)) \] \[ = \mathbf{i}(-1 + 4) - \mathbf{j}(-2 - 4) + \mathbf{k}(-4 - 2) \] \[ = 3\mathbf{i} + 6\mathbf{j} - 6\mathbf{k} \] Now, find the magnitude of this cross product:
\[ |\vec{AP} \times \vec{b}| = \sqrt{3^2 + 6^2 + (-6)^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9 \] Next, find the magnitude of the direction vector \(\vec{b}\):
\[ |\vec{b}| = \sqrt{(-2)^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \] Finally, calculate the shortest distance \(d\):
\[ d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|} = \frac{9}{3} = 3 \] Step 4: Final Answer:
The shortest distance between the point and the line is 3. This corresponds to option (C).