Question

The length of perpendicular drawn from the point (3, -1, 11) to the line \(\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) is

• A. \(\sqrt{33}\)
• B. \(\sqrt{66}\)
• C. \(\sqrt{53}\)
• D. \(\sqrt{29}\)

Hint:

To find the perpendicular distance from a point to a line in 3D, use the formula \( d = \frac{| \vec{P} \times \vec{d} |}{|\vec{d}|} \), where \( \vec{P} \) is the vector from the point to the line and \( \vec{d} \) is the direction vector of the line.

Answer 1

The Correct Option is C

The correct answer is: (C): \(\sqrt{53}\)

We are given the point \( P(3, -1, 11) \) and the line equation:

\(\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}\)

Step 1: Parametrize the line equation

We start by parametrizing the given line equation. Let \( \lambda \) be the parameter. The parametric form of the line is:

\( x = 2\lambda, \quad y = 3\lambda + 2, \quad z = 4\lambda + 3 \)

Step 2: Find the direction vector of the line

The direction vector \( \vec{d} \) of the line is the vector of coefficients of \( \lambda \), which is:

\( \vec{d} = (2, 3, 4) \)

Step 3: Find the vector from the point to the line

The point on the line corresponding to the parameter \( \lambda \) is \( (2\lambda, 3\lambda + 2, 4\lambda + 3) \). The vector from this point to \( P(3, -1, 11) \) is:

\( \vec{P} = (3 - 2\lambda, -1 - (3\lambda + 2), 11 - (4\lambda + 3)) \)

Simplifying the vector components:

\( \vec{P} = (3 - 2\lambda, -3 - 3\lambda, 8 - 4\lambda) \)

Step 4: Use the formula for the perpendicular distance

The perpendicular distance \( d \) from a point to a line is given by the formula:

\( d = \frac{| \vec{P} \times \vec{d} |}{|\vec{d}|} \)

Step 5: Compute the cross product

We first compute the cross product \( \vec{P} \times \vec{d} \). We have:

\( \vec{P} = (3 - 2\lambda, -3 - 3\lambda, 8 - 4\lambda), \quad \vec{d} = (2, 3, 4) \)

Using the determinant formula for the cross product, we calculate:

\( \vec{P} \times \vec{d} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 - 2\lambda & -3 - 3\lambda & 8 - 4\lambda \\ 2 & 3 & 4 \end{matrix} \right| \)

After evaluating the determinant, we get:

\( \vec{P} \times \vec{d} = (-12 + 8\lambda, 6 - 4\lambda, 9 - 6\lambda) \)

Step 6: Find the magnitude of the cross product

Now, we find the magnitude of \( \vec{P} \times \vec{d} \):

\( | \vec{P} \times \vec{d} | = \sqrt{(-12 + 8\lambda)^2 + (6 - 4\lambda)^2 + (9 - 6\lambda)^2} \)

Step 7: Solve for the perpendicular distance

Finally, we divide the magnitude of the cross product by the magnitude of \( \vec{d} \), which is:

\( |\vec{d}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29} \)

After simplifying the expression, we find that the perpendicular distance is:

\( d = \sqrt{53} \)

Conclusion:
The length of the perpendicular drawn from the point \( (3, -1, 11) \) to the line \( \frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \) is \( \sqrt{53} \), so the correct answer is (C): \(\sqrt{53}\).