Question

Let \((\alpha,\beta,\gamma)\) be the foot of the perpendicular from the point \((25,2,41)\) on the line \[ \frac{x-4}{3}=\frac{y+1}{7}=\frac{z-2}{3}. \] Then \(\alpha+\beta+\gamma\) is equal to:

• A. \(41\)
• B. \(42\)
• C. \(44\)
• D. \(45\)

Hint:

For foot of perpendicular problems in 3D, always use the dot product condition with the direction vector of the line.

Answer 1

The Correct Option is C

We are required to find the value of \[ \alpha + \beta + \gamma, \] where \((\alpha, \beta, \gamma)\) is the foot of the perpendicular from the point \((25,\,2,\,41)\) to the given line.

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Step 1: Equation of the line

The line is given in symmetric form:

\[ \frac{x-4}{3} = \frac{y+1}{7} = \frac{z-2}{3} = t \]

Hence, its parametric equations are:

• \(x = 3t + 4\)
• \(y = 7t - 1\)
• \(z = 3t + 2\)

The direction ratios of the line are: \[ (3,\,7,\,3). \]

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Step 2: Condition for foot of the perpendicular

Let the foot of the perpendicular be \((\alpha, \beta, \gamma)\). Then,

\[ (\alpha, \beta, \gamma) = (3t+4,\; 7t-1,\; 3t+2). \]

The vector from \((\alpha, \beta, \gamma)\) to \((25, 2, 41)\) is:

\[ (25-\alpha,\; 2-\beta,\; 41-\gamma). \]

Since this vector is perpendicular to the direction vector \((3,7,3)\), their dot product must be zero:

\[ 3(25-\alpha) + 7(2-\beta) + 3(41-\gamma) = 0. \]

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Step 3: Simplification

Expanding the dot product:

\[ 75 - 3\alpha + 14 - 7\beta + 123 - 3\gamma = 0 \]

Combining terms:

\[ 212 - (3\alpha + 7\beta + 3\gamma) = 0 \]

\[ \Rightarrow \quad 3\alpha + 7\beta + 3\gamma = 212. \]

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Step 4: Use parametric values

Substitute:

• \(\alpha = 3t + 4\)
• \(\beta = 7t - 1\)
• \(\gamma = 3t + 2\)

Then,

\[ \alpha + \beta + \gamma = (3t+4) + (7t-1) + (3t+2) = 13t + 5. \]

From the perpendicularity condition:

\[ 3\alpha + 7\beta + 3\gamma = 212 \;\Rightarrow\; 13t + 5 = 44. \]

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Final Answer

\[ \boxed{44} \]