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: Understanding the Concept:
We first need to find the direction of the required line. Since the line is perpendicular to two given vectors, its direction vector will be given by their cross product.

Step 2: Key Formula or Approach:
Direction vector: \[ \vec{d} = (\hat{i} + 2\hat{j} + 2\hat{k}) \times (2\hat{i} + 2\hat{j} + \hat{k}) \] Line equation: \[ \vec{r} = \vec{a} + \lambda \vec{d} \]

Step 3: Detailed Explanation:
Compute the cross product: \[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} \] \[ = \hat{i}(2 \cdot 1 - 2 \cdot 2) - \hat{j}(1 \cdot 1 - 2 \cdot 2) + \hat{k}(1 \cdot 2 - 2 \cdot 2) \] \[ = \hat{i}(2 - 4) - \hat{j}(1 - 4) + \hat{k}(2 - 4) = -2\hat{i} + 3\hat{j} - 2\hat{k} \] Equation of line passing through \( (1,1,1) \): \[ x = 1 - 2\lambda,\quad y = 1 + 3\lambda,\quad z = 1 - 2\lambda \] Foot of perpendicular from origin satisfies: \[ \vec{r} \cdot \vec{d} = 0 \] Substituting: \[ (-2)(1 - 2\lambda) + 3(1 + 3\lambda) + (-2)(1 - 2\lambda) = 0 \] \[ -2 + 4\lambda + 3 + 9\lambda - 2 + 4\lambda = 0 \] \[ 17\lambda - 1 = 0 \Rightarrow \lambda = \frac{1}{17} \] Coordinates of foot: \[ a = 1 - \frac{2}{17} = \frac{15}{17},\quad b = 1 + \frac{3}{17} = \frac{20}{17},\quad c = 1 - \frac{2}{17} = \frac{15}{17} \] Sum: \[ a + b + c = \frac{15 + 20 + 15}{17} = \frac{50}{17} \] Required value: \[ 34(a + b + c) = 34 \times \frac{50}{17} = 2 \times 50 = 100 \]

Step 4: Final Answer:
100