Question

A(1, -2, 1) and B(2, -1, 2) are the end points of a line segment. If \(D(\alpha, \beta, \gamma)\) is the foot of the perpendicular drawn from \(C(1, 2, 3)\) to AB, then \(\alpha^2 + \beta^2 + \gamma^2 =\)

• A. \(\sqrt{18}\)
• B. 14
• C. 9
• D. 27

Hint:

- The projection formula is very useful in geometry for finding specific points such as the foot of the perpendicular. In this case, it allowed us to determine the coordinates of \(D\).

Answer 1

The Correct Option is A

Step 1: Find the direction vector of line AB. 
- The direction vector \(\vec{AB}\) is found by subtracting the coordinates of point \(A(1, -2, 1)\) from point \(B(2, -1, 2)\): \[ \vec{AB} = B - A = (2 - 1, -1 - (-2), 2 - 1) = (1, 1, 1) \]

 Step 2: Find the vector \(\vec{AC}\). 
- The vector \(\vec{AC}\) is found by subtracting the coordinates of point \(A(1, -2, 1)\) from point \(C(1, 2, 3)\): \[ \vec{AC} = C - A = (1 - 1, 2 - (-2), 3 - 1) = (0, 4, 2) \] 

Step 3: Find the projection of \(\vec{AC}\) onto \(\vec{AB}\). 
- The formula for the projection of vector \(\vec{AC}\) onto vector \(\vec{AB}\) is: \[ \text{proj}_{\vec{AB}} \vec{AC} = \frac{\vec{AC} \cdot \vec{AB}}{\vec{AB} \cdot \vec{AB}} \vec{AB} \] First, calculate the dot product \(\vec{AC} \cdot \vec{AB}\): \[ \vec{AC} \cdot \vec{AB} = (0)(1) + (4)(1) + (2)(1) = 0 + 4 + 2 = 6 \] Next, calculate \(\vec{AB} \cdot \vec{AB}\): \[ \vec{AB} \cdot \vec{AB} = (1)(1) + (1)(1) + (1)(1) = 1 + 1 + 1 = 3 \] Now, apply the projection formula: \[ \text{proj}_{\vec{AB}} \vec{AC} = \frac{6}{3} \vec{AB} = 2 \cdot (1, 1, 1) = (2, 2, 2) \]

 Step 4: Find the coordinates of point D, the foot of the perpendicular. 
- The coordinates of \(D(\alpha, \beta, \gamma)\) are found by subtracting the projection vector from point \(C\): \[ \vec{CD} = \vec{AC} - \text{proj}_{\vec{AB}} \vec{AC} = (0, 4, 2) - (2, 2, 2) = (-2, 2, 0) \] Thus, the coordinates of point \(D\) are \((-2, 2, 0)\), so \(\alpha = -2\), \(\beta = 2\), and \(\gamma = 0\). 

 Step 5: Calculate \(\alpha^2 + \beta^2 + \gamma^2\). 
- Finally, calculate: \[ \alpha^2 + \beta^2 + \gamma^2 = (-2)^2 + 2^2 + 0^2 = 4 + 4 + 0 = 8 \] Thus, \(\alpha^2 + \beta^2 + \gamma^2 = 8\).