Question

Foot of perpendicular from origin on a line passing through $(1, 1, 1)$ having direction ratios $\langle 2, 3, 4 \rangle$, is:

• A. $(\frac{11}{29}, \frac{2}{29}, \frac{7}{29})$
• B. $(\frac{11}{29}, \frac{-2}{29}, \frac{-7}{29})$
• C. $(\frac{11}{29}, \frac{2}{29}, \frac{-7}{29})$
• D. $(\frac{-11}{29}, \frac{2}{29}, \frac{-7}{29})$

Answer 1

The Correct Option is C

Step 1: Understanding the Concept: 
The problem asks for the foot of the perpendicular \( N \) from the origin \( O(0,0,0) \) to a specific line. 
The line passes through a point \( A(1, 1, 1) \) and has direction ratios \( \vec{d} = \langle 2, 3, 4 \rangle \). 
We know that the vector \( \vec{ON} \) must be perpendicular to the line's direction vector. 
Step 2: Key Formula or Approach: 
The parametric equation of a line passing through \( (x_1, y_1, z_1) \) with direction ratios \( a, b, c \) is: 
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda \] 
The condition for perpendicularity between two vectors \( \vec{u} \) and \( \vec{v} \) is \( \vec{u} \cdot \vec{v} = 0 \). 
Step 3: Detailed Explanation: 
Any point \( N \) on the line can be represented as: 
\[ x = 2\lambda + 1, \quad y = 3\lambda + 1, \quad z = 4\lambda + 1 \] 
The vector \( \vec{ON} \) from the origin to this point is \( (2\lambda + 1)\hat{i} + (3\lambda + 1)\hat{j} + (4\lambda + 1)\hat{k} \). 
Since \( \vec{ON} \) is perpendicular to the line (direction vector \( 2\hat{i} + 3\hat{j} + 4\hat{k} \)): 
\[ 2(2\lambda + 1) + 3(3\lambda + 1) + 4(4\lambda + 1) = 0 \] 
\[ 4\lambda + 2 + 9\lambda + 3 + 16\lambda + 4 = 0 \] 
\[ 29\lambda + 9 = 0 \implies \lambda = -\frac{9}{29} \] 
Substitute \( \lambda \) back into the parametric coordinates: 
\( x = 2\left(-\frac{9}{29}\right) + 1 = \frac{-18 + 29}{29} = \frac{11}{29} \) 
\( y = 3\left(-\frac{9}{29}\right) + 1 = \frac{-27 + 29}{29} = \frac{2}{29} \) 
\( z = 4\left(-\frac{9}{29}\right) + 1 = \frac{-36 + 29}{29} = -\frac{7}{29} \) 
Step 4: Final Answer: 
The coordinates of the foot of the perpendicular are \( \left( \frac{11}{29}, \frac{2}{29}, -\frac{7}{29} \right) \).