Question

A line through $(1, 1, 1)$ and perpendicular to both $\hat{i} + 2\hat{j} + 2\hat{k}$ and $2\hat{i} + 2\hat{j} + \hat{k}$, let $(a, b, c)$ be foot of perpendicular from origin then $34 (a + b + c)$ is:

Answer 1

Step 1: Find the direction vector of the line.
The line is perpendicular to $\u0000u = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\u0000v = 2\hat{i} + 2\hat{j} + \hat{k}$.
Its direction vector $\u0000d = \u0000u \times \u0000v = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix}$
$\u0000d = \hat{i}(2-4) - \hat{j}(1-4) + \hat{k}(2-4) = -2\hat{i} + 3\hat{j} - 2\hat{k}$.


Step 2: Write the equation of the line.
Line passes through $(1, 1, 1)$ with direction $(-2, 3, -2)$.
Equation: $\frac{x-1}{-2} = \frac{y-1}{3} = \frac{z-1}{-2} = \lambda$.
General point $M = (-2\lambda+1, 3\lambda+1, -2\lambda+1)$.


Step 3: Find the foot of perpendicular from origin.
If $M(a, b, c)$ is the foot of the perpendicular from $O(0,0,0)$, then $\vec{OM} \cdot \u0000d = 0$.
$(-2\lambda+1)(-2) + (3\lambda+1)(3) + (-2\lambda+1)(-2) = 0$
$4\lambda - 2 + 9\lambda + 3 + 4\lambda - 2 = 0$
$17\lambda - 1 = 0 \implies \lambda = \frac{1}{17}$


Step 4: Calculate $a+b+c$.
$a = -2(1/17) + 1 = 15/17$
$b = 3(1/17) + 1 = 20/17$
$c = -2(1/17) + 1 = 15/17$
$a + b + c = \frac{15 + 20 + 15}{17} = \frac{50}{17}$


Step 5: Final value.
$34 (a + b + c) = 34 \cdot \frac{50}{17} = 2 \cdot 50 = 100$.
The answer is 100.