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: Identify a point on the line and its direction vector.
The given line is \[ \frac{x-4}{-2}=\frac{y-4}{2}=\frac{z-6}{1} \] Comparing with the symmetric form \[ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \] we get a point on the line as \[ A=(4,4,6) \] and the direction vector of the line as \[ \vec{d}=(-2,2,1) \]

Step 2: Write the coordinates of the given external point.
The given point is \[ P=(2,3,4) \] We now form the vector from the point on the line \( A \) to the external point \( P \): \[ \overrightarrow{AP}=P-A=(2-4,\ 3-4,\ 4-6) \] \[ \overrightarrow{AP}=(-2,-1,-2) \]

Step 3: Recall the formula for shortest distance from a point to a line in 3D.
If \( \vec{r} \) is the vector from a point on the line to the given point, and \( \vec{d} \) is the direction vector of the line, then the shortest distance is \[ \text{Distance}=\frac{|\vec{r}\times\vec{d}|}{|\vec{d}|} \] Here, \[ \vec{r}=\overrightarrow{AP}=(-2,-1,-2) \quad \text{and} \quad \vec{d}=(-2,2,1) \]

Step 4: Find the cross product \( \overrightarrow{AP}\times\vec{d} \).
\[ \overrightarrow{AP}\times\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} -2 & -1 & -2 -2 & 2 & 1 \end{vmatrix} \] Expanding, \[ = \hat{i}\bigl((-1)(1)-(-2)(2)\bigr) -\hat{j}\bigl((-2)(1)-(-2)(-2)\bigr) +\hat{k}\bigl((-2)(2)-(-1)(-2)\bigr) \] \[ = \hat{i}(-1+4)-\hat{j}(-2-4)+\hat{k}(-4-2) \] \[ =3\hat{i}+6\hat{j}-6\hat{k} \]

Step 5: Find the magnitudes needed in the formula.
Now, \[ |\overrightarrow{AP}\times\vec{d}|=\sqrt{3^2+6^2+(-6)^2} \] \[ =\sqrt{9+36+36}=\sqrt{81}=9 \] Also, \[ |\vec{d}|=\sqrt{(-2)^2+2^2+1^2} \] \[ =\sqrt{4+4+1}=\sqrt{9}=3 \]

Step 6: Apply the distance formula.
Therefore, \[ \text{Shortest distance}=\frac{|\overrightarrow{AP}\times\vec{d}|}{|\vec{d}|} =\frac{9}{3}=3 \]

Step 7: Final conclusion.
Hence, the shortest distance between the point and the line is \[ \boxed{3} \] Therefore, the correct option is \[ \boxed{(3)\ 3} \]