AP EAMCET · Maths · Three Dimensional Geometry
On a line with direction cosines \(l, \mathrm{~m}, \mathrm{n}, \mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) is a fixed point. If \(\mathrm{B}=\left(\mathrm{x}_1+4 \mathrm{k} l, \mathrm{y}_1+4 \mathrm{~km}, \mathrm{z}_1+4 \mathrm{kn}\right)\) and \(\mathrm{C}=\left(\mathrm{x}_1+\mathrm{k} l, \mathrm{y}_1+\mathrm{km}, \mathrm{z}_1+\mathrm{kn}\right)(\mathrm{k}>0)\) then the ratio in which the point B divides the line segment joining A and C is
- A \(1: 2\)
- B \(1:-4\)
- C \(4:-3\)
- D \(4: 3\)
Answer & Solution
Correct Answer
(C) \(4:-3\)
Step-by-step Solution
Detailed explanation
\( \vec{A} = (x_1, y_1, z_1) \) \( \vec{C} = (x_1+kl, y_1+km, z_1+kn) \) \( \vec{B} = (x_1+4kl, y_1+4km, z_1+4kn) \) Let B divide AC in ratio \( \lambda:1 \). \( x_B = \frac{\lambda x_C + 1 \cdot x_A}{\lambda+1} \) \( x_1+4kl = \frac{\lambda(x_1+kl) + x_1}{\lambda+1} \)…
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