Question

Let \(\alpha, \beta, \gamma\) be the foot of perpendicular from the point \((1, 2, 3)\) on the line \(\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}\). Then \(19(\alpha + \beta + \gamma)\) is equal to:

• A. 102
• B. 101
• C. 99
• D. 100

Answer 1

The Correct Option is B

Given the line \(\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}\), we can parametrize it as:
\[x = 5t - 3, \quad y = 2t + 1, \quad z = 3t - 4\]

Let \(P(\alpha, \beta, \gamma) = (5t - 3, 2t + 1, 3t - 4)\) be the foot of the perpendicular from \(A = (1, 2, 3)\) to the line. The vector \(\overrightarrow{AP}\) is:

\[\overrightarrow{AP} = (5t - 4, 2t - 1, 3t - 7)\]

Since \(\overrightarrow{AP}\) is perpendicular to the line, we set up the dot product with the direction ratios \((5, 2, 3)\):

\[(5t - 4) \times 5 + (2t - 1) \times 2 + (3t - 7) \times 3 = 0\]

Expanding and solving:

\[38t - 43 = 0 \Rightarrow t = \frac{43}{38}\]

Substitute \(t = \frac{43}{38}\) to find \(\alpha\), \(\beta\), and \(\gamma\):

\[\alpha = 5t - 3 = \frac{101}{38}, \quad \beta = 2t + 1 = \frac{62}{19}, \quad \gamma = 3t - 4 = \frac{-23}{38}\]

Then,

\[\alpha + \beta + \gamma = \frac{101}{38} + \frac{124}{38} - \frac{23}{38} = \frac{202}{38} = \frac{101}{19}\]

Finally,

\[19(\alpha + \beta + \gamma) = 101\]

Answer 2

To find the value of \(19(\alpha + \beta + \gamma)\), we need to determine the coordinates of the foot of the perpendicular, \((\alpha, \beta, \gamma)\), from the point \((1, 2, 3)\) to the given line. The line is given in symmetric form as:

\(\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}\)

The direction ratios of the line are \(5, 2, \text{ and } 3\). Let's denote the point on the line as \((x_1, y_1, z_1)\), parameterized as:

• \(x_1 = -3 + 5t\)
• \(y_1 = 1 + 2t\)
• \(z_1 = -4 + 3t\)

The coordinates of the foot of the perpendicular can be found using the formula for the line joining the point \((x_0, y_0, z_0)\) and any point on the line \((x_1, y_1, z_1)\). The direction ratios of the foot of the perpendicular from \((x_0, y_0, z_0)\) are along the vector connecting these points and perpendicular to the given line's direction ratios \(5, 2, 3\). Thus:

\(5(x_1 - 1) + 2(y_1 - 2) + 3(z_1 - 3) = 0\)

Substituting the parameterized coordinates:

• \(x_1 - 1 = -4 + 5t\)
• \(y_1 - 2 = -1 + 2t\)
• \(z_1 - 3 = -7 + 3t\)

The equation becomes:

\(5(-4 + 5t) + 2(-1 + 2t) + 3(-7 + 3t) = 0\)

Simplifying this gives:

\(-20 + 25t - 2 + 4t - 21 + 9t = 0\)

\(38t - 43 = 0\)

\(t = \frac{43}{38}\)

Substitute \(t\) back into the parameterized equations:

• \(\alpha = -3 + 5\frac{43}{38} = \frac{-114 + 215}{38} = \frac{101}{38}\)
• \(\beta = 1 + 2\frac{43}{38} = \frac{38 + 86}{38} = \frac{124}{38}\)
• \(\gamma = -4 + 3\frac{43}{38} = \frac{-152 + 129}{38} = \frac{-23}{38}\)

The sum of these coordinates is:

\(\alpha + \beta + \gamma = \frac{101}{38} + \frac{124}{38} - \frac{23}{38} = \frac{202}{38} = \frac{101}{19}\)

Finally, the value of \(19(\alpha + \beta + \gamma)\) is:

\(19 \times \frac{101}{19} = 101\)

Thus, the correct answer is \(101\).