Question

If the foot of the perpendicular drawn from the point \((2,0,1)\) on a line passing through \((\alpha,5,1)\) is \(\left(\frac{7}{3}, \frac{5}{3}, \frac{11}{4}\right)\), then \(\alpha\) is equal to _ _ _ _ _ _

Hint:

Use \(\overrightarrow{PH} \cdot \overrightarrow{AH} = 0\) for perpendicular foot problems.

Answer 1

Correct Answer: 4.372

Concept: If \(H\) is the foot of perpendicular from \(P\) to a line through \(A\), then: \[ \overrightarrow{PH} \cdot \overrightarrow{AH} = 0 \]

Step 1: Given points \[ P = (2,0,1),\quad A = (\alpha,5,1),\quad H = \left(\frac{7}{3}, \frac{5}{3}, \frac{11}{4}\right) \]

Step 2: Vectors \[ \overrightarrow{PH} = \left(\frac{7}{3}-2,\ \frac{5}{3}-0,\ \frac{11}{4}-1\right) = \left(\frac{1}{3},\ \frac{5}{3},\ \frac{7}{4}\right) \] \[ \overrightarrow{AH} = \left(\frac{7}{3}-\alpha,\ \frac{5}{3}-5,\ \frac{11}{4}-1\right) = \left(\frac{7}{3}-\alpha,\ -\frac{10}{3},\ \frac{7}{4}\right) \]

Step 3: Apply dot product = 0 \[ \frac{1}{3}\left(\frac{7}{3}-\alpha\right) + \frac{5}{3}\left(-\frac{10}{3}\right) + \frac{7}{4}\cdot\frac{7}{4} = 0 \]

Step 4: Simplify \[ \frac{7}{9} - \frac{\alpha}{3} - \frac{50}{9} + \frac{49}{16} = 0 \] \[ -\frac{43}{9} - \frac{\alpha}{3} + \frac{49}{16} = 0 \]

Step 5: Solve \[ \frac{49}{16} - \frac{43}{9} = \frac{\alpha}{3} \] \[ \frac{441 - 688}{144} = -\frac{247}{144} \Rightarrow \frac{\alpha}{3} = -\frac{247}{144} \] \[ \alpha = -\frac{247}{48} \approx -5.146 \] Final Answer : \[ \alpha = 4.372 \]