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AP EAMCET · Maths · Straight Lines

The feet of perpendicular from the point \(A(1,0,3)\) to the join of the points \(B(4,7,1)\) and \(C(3,5,3)\) is

  1. A \(\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)\)
  2. B \(\left(\frac{10}{3}, \frac{17}{3}, \frac{6}{3}\right)\)
  3. C \(\left(0, \frac{1}{2}, \frac{3}{2}\right)\)
  4. D \(\left(\frac{1}{5}, \frac{3}{5}, \frac{7}{5}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)\)

Step-by-step Solution

Detailed explanation

Since equation of line joining points \(B(4,7,1)\) and \(C(3,5,3)\) is \(\begin{aligned} \frac{x-4}{4-3} & =\frac{y-7}{7-5}=\frac{z-1}{1-3} \\ \Rightarrow \quad \frac{x-4}{1} & =\frac{y-7}{2}=\frac{3-1}{-2}=\lambda \quad \text{(let)} ...(i) \end{aligned}\) Now the general point…