Question

If \( 3\overline{i} + \overline{j} + \overline{k} \), \( 2\overline{i} + \overline{k} \), and \( \overline{i} + 5\overline{j} \) are the position vectors of three non-collinear points A, B, C respectively. If the perpendicular drawn from C onto \( \overline{AB} \) meets \( \overline{AB} \) at the point \( a\overline{i} + b\overline{j} + c\overline{k} \), then \( a + b + c = \)

• A. $ 5 $
• B. $ 3 $
• C. $ 7 $
• D. $ 9 $

Hint:

To find the foot of the perpendicular from a point to a line, use the projection formula to determine the component of the vector along the line and add it to the starting point of the line.

Answer 1

The Correct Option is C

Step 1: Define the position vectors. 
Let: $$ \overrightarrow{OA} = 3\overline{i} + \overline{j} + \overline{k}, \quad \overrightarrow{OB} = 2\overline{i} + \overline{k}, \quad \overrightarrow{OC} = \overline{i} + 5\overline{j}. $$ The vector \( \overline{AB} \) is: $$ \overline{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2\overline{i} + \overline{k}) - (3\overline{i} + \overline{j} + \overline{k}) = -\overline{i} - \overline{j}. $$ The vector \( \overline{AC} \) is: $$ \overline{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (\overline{i} + 5\overline{j}) - (3\overline{i} + \overline{j} + \overline{k}) = -2\overline{i} + 4\overline{j} - \overline{k}. $$ Step 2: Find the projection of \( \overline{AC} \) onto \( \overline{AB} \). 
The projection of \( \overline{AC} \) onto \( \overline{AB} \) is given by: $$ \text{proj}_{\overline{AB}} \overline{AC} = \frac{\overline{AC} \cdot \overline{AB}}{\overline{AB} \cdot \overline{AB}} \cdot \overline{AB}. $$ First, compute \( \overline{AC} \cdot \overline{AB} \): $$ \overline{AC} \cdot \overline{AB} = (-2\overline{i} + 4\overline{j} - \overline{k}) \cdot (-\overline{i} - \overline{j}) = (-2)(-1) + (4)(-1) + (-1)(0) = 2 - 4 + 0 = -2. $$ Next, compute \( \overline{AB} \cdot \overline{AB} \): $$ \overline{AB} \cdot \overline{AB} = (-\overline{i} - \overline{j}) \cdot (-\overline{i} - \overline{j}) = (-1)^2 + (-1)^2 = 1 + 1 = 2. $$ Thus: $$ \text{proj}_{\overline{AB}} \overline{AC} = \frac{-2}{2} \cdot (-\overline{i} - \overline{j}) = -1 \cdot (-\overline{i} - \overline{j}) = \overline{i} + \overline{j}. $$ Step 3: Find the coordinates of the foot of the perpendicular. 
The foot of the perpendicular from \( C \) to \( \overline{AB} \) is: $$ \overrightarrow{OP} = \overrightarrow{OA} + \text{proj}_{\overline{AB}} \overline{AC}. $$ Substitute: $$ \overrightarrow{OP} = (3\overline{i} + \overline{j} + \overline{k}) + (\overline{i} + \overline{j}) = 4\overline{i} + 2\overline{j} + \overline{k}. $$ Thus, the coordinates of the foot of the perpendicular are \( (4, 2, 1) \). 
Step 4: Compute \( a + b + c \). 
Here, \( a = 4 \), \( b = 2 \), and \( c = 1 \). Therefore: $$ a + b + c = 4 + 2 + 1 = 7. $$ Step 5: Final Answer. 
$$ \boxed{7} $$