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Step-by-Step Solution
Step 1: Identify the Relevant Points and Vectors
• The given point from which the perpendicular is drawn is
(1,\,0,\,3) .
• The line passes through (\alpha,\,7,\,1) .
• The foot of the perpendicular on this line is
\left(\frac{5}{3},\,\frac{7}{3},\,\frac{17}{3}\right).
Step 2: Express the Direction Vector of the Line
The direction vector of the line can be taken as the vector
from (\alpha,\,7,\,1) to
\left(\frac{5}{3},\,\frac{7}{3},\,\frac{17}{3}\right) :
\overrightarrow{d}
=
\left(\frac{5}{3} - \alpha,\;\frac{7}{3} - 7,\;\frac{17}{3} - 1\right)
=
\left(\frac{5}{3} - \alpha,\;\frac{7}{3} - \frac{21}{3},\;\frac{17}{3} - \frac{3}{3}\right)
=
\left(\frac{5}{3} - \alpha,\;-\frac{14}{3},\;\frac{14}{3}\right).
Step 3: Form the Vector from the External Point to the Foot
The vector from (1,\,0,\,3) to the foot
\left(\frac{5}{3},\,\frac{7}{3},\,\frac{17}{3}\right) is:
\overrightarrow{PF}
=
\left(\frac{5}{3} - 1,\;\frac{7}{3} - 0,\;\frac{17}{3} - 3\right)
=
\left(\frac{2}{3},\;\frac{7}{3},\;\frac{8}{3}\right).
Step 4: Use the Perpendicularity Condition (Dot Product = 0)
Since \overrightarrow{PF} is perpendicular to the direction vector
\overrightarrow{d} , their dot product must be zero:
\left(\frac{5}{3} - \alpha\right)\left(\frac{2}{3}\right)
\;+\;
\left(-\frac{14}{3}\right)\left(\frac{7}{3}\right)
\;+\;
\left(\frac{14}{3}\right)\left(\frac{8}{3}\right)
= 0.
Step 5: Simplify and Solve for \alpha
Simplify each term:
\frac{2}{3}\left(\frac{5}{3} - \alpha\right)
= \frac{2\,(5 - 3\alpha)}{9},
\quad
\left(-\frac{14}{3}\right)\left(\frac{7}{3}\right)
= -\frac{98}{9},
\quad
\left(\frac{14}{3}\right)\left(\frac{8}{3}\right)
= \frac{112}{9}.
Putting these together gives:
\frac{2(5 - 3\alpha)}{9}
- \frac{98}{9}
+ \frac{112}{9}
= 0.
Combine the numeric terms:
- \frac{98}{9} + \frac{112}{9} = \frac{14}{9},
so
\frac{2(5 - 3\alpha)}{9} + \frac{14}{9} = 0.
Multiply through by 9 to clear the denominator:
2(5 - 3\alpha) + 14 = 0.
Simplify:
10 - 6\alpha + 14 = 0
\;\Longrightarrow\;
24 - 6\alpha = 0
\;\Longrightarrow\;
6\alpha = 24
\;\Longrightarrow\;
\alpha = 4.
Final Answer
\alpha = 4.
Reference Image
