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: Write the vector equation of the line.
Line passes through $A(1, 1, 1)$ with direction vector $\u0000d = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
$\u0000r = (1, 1, 1) + \lambda(2, 3, 4)$.
A general point $N$ on this line is $(2\lambda + 1, 3\lambda + 1, 4\lambda + 1)$.

Step 2: Use the perpendicular condition.
If $N$ is the foot of the perpendicular from the origin $O(0, 0, 0)$, then the vector $\vec{ON}$ must be perpendicular to the line's direction vector $\u0000d$.
$\vec{ON} = (2\lambda + 1)\hat{i} + (3\lambda + 1)\hat{j} + (4\lambda + 1)\hat{k}$.
Condition: $\vec{ON} \cdot \u0000d = 0$.
$2(2\lambda + 1) + 3(3\lambda + 1) + 4(4\lambda + 1) = 0$
$4\lambda + 2 + 9\lambda + 3 + 16\lambda + 4 = 0$
$29\lambda + 9 = 0 \implies \lambda = -\frac{9}{29}$

Step 3: Find the coordinates of $N$.
$x = 2(-\frac{9}{29}) + 1 = -\frac{18}{29} + \frac{29}{29} = \frac{11}{29}$
$y = 3(-\frac{9}{29}) + 1 = -\frac{27}{29} + \frac{29}{29} = \frac{2}{29}$
$z = 4(-\frac{9}{29}) + 1 = -\frac{36}{29} + \frac{29}{29} = -\frac{7}{29}$
The foot is $(\frac{11}{29}, \frac{2}{29}, -\frac{7}{29})$, matching option (3).