Question

On a line with direction cosines l, m, n, \( A(x_1, y_1, z_1) \) is a fixed point. If \( B=(x_1+4kl, y_1+4km, z_1+4kn) \) and \( C=(x_1+kl, y_1+km, z_1+kn) \) (\(k>0\)) 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

Hint:

If a point R divides segment PQ in ratio \(m:n\), then \( R = \frac{n\vec{P} + m\vec{Q}}{m+n} \). A negative ratio indicates external division. Consider the points on a line. Let A be the origin (0). Then \(C\) is at a position vector \(k(l,m,n)\) and \(B\) is at \(4k(l,m,n)\). So the coordinates are \(A \leftrightarrow 0\), \(C \leftrightarrow k\), \(B \leftrightarrow 4k\). If B divides AC in ratio \( \lambda:1 \), then \( 4k = \frac{\lambda \cdot k + 1 \cdot 0}{\lambda+1} \). \( 4(\lambda+1) = \lambda \implies 4\lambda+4=\lambda \implies 3\lambda = -4 \implies \lambda = -4/3 \). The ratio is \(-4/3 : 1\), which is \(-4:3\). This corresponds to \(m=4, n=-3\) or \(m=-4, n=3\).

Answer 1

The Correct Option is C

Let \( A, B, C \) be points on a line.
\( A = (x_1, y_1, z_1) \)
\( B = (x_1 + 4kl, y_1 + 4km, z_1 + 4kn) \)
\( C = (x_1 + kl, y_1 + km, z_1 + kn) \)
Let \( B \) divide \( AC \) in the ratio \( \lambda : 1 \).
Using the section formula for the x-coordinate: \[ x_B = \frac{\lambda x_C + 1 x_A}{\lambda + 1} \] \[ x_1 + 4kl = \frac{\lambda (x_1 + kl) + 1 (x_1)}{\lambda + 1} \] \[ (\lambda + 1)(x_1 + 4kl) = \lambda x_1 + \lambda kl + x_1 \] \[ \lambda x_1 + 4\lambda kl + x_1 + 4kl = \lambda x_1 + \lambda kl + x_1 \] Subtract \( \lambda x_1 + x_1 \) from both sides: \[ 4\lambda kl + 4kl = \lambda kl \] Since \( k > 0 \) and \( l, m, n \) are direction cosines (not all zero), we can assume \( kl \neq 0 \) (unless \( l = 0 \), but the formula must hold for \( y \) and \( z \) coordinates too).
If \( kl \neq 0 \), divide by \( kl \): \[ 4\lambda + 4 = \lambda \] \[ 3\lambda = -4 \] \[ \lambda = -\frac{4}{3} \] The ratio \( \lambda : 1 \) is \( -\frac{4}{3} : 1 \), which is equivalent to \( -4 : 3 \).
A negative ratio means \( B \) divides \( AC \) externally.
If the ratio is taken as \( m : n \), then \( \frac{m}{n} = -\frac{4}{3} \).
This can be written as \( m = 4, n = -3 \) or \( m = -4, n = 3 \).
The option \( 4 : -3 \) corresponds to \( m = 4, n = -3 \).
This means \( B \) divides \( AC \) externally in the ratio 4:3, and \( C \) is between \( A \) and \( B \).
Let's check the distances.
Distance \( AB = \sqrt{((x_1 + 4kl) - x_1)^2 + \dots} = \sqrt{(4kl)^2 + (4km)^2 + (4kn)^2} = \sqrt{16k^2(l^2 + m^2 + n^2)} = \sqrt{16k^2(1)} = 4k \) (since \( k > 0 \)).
Distance \( AC = \sqrt{((x_1 + kl) - x_1)^2 + \dots} = \sqrt{(kl)^2 + (km)^2 + (kn)^2} = \sqrt{k^2(l^2 + m^2 + n^2)} = \sqrt{k^2(1)} = k \).
The points are ordered \( A, C, B \) along the line if \( k > 0 \).
\( A \) is the origin, \( C \) is at distance \( k \), \( B \) is at distance \( 4k \).
\( A \)-------\( C \)---\( B \)
\( AC = k, CB = 3k \).
\( B \) divides \( AC \) externally.
Ratio \( AB / BC \).
This would be \( \frac{4k}{-3k} = -\frac{4}{3} \).
The ratio in which \( B \) divides segment \( AC \) is given by \( \frac{\vec{AB}}{\vec{BC}} \).
\( \vec{AC} = (kl, km, kn) \). \( \vec{AB} = (4kl, 4km, 4kn) = 4 \vec{AC} \).
So \( B \) is such that \( \vec{OB} = \vec{OA} + 4(\vec{OC} - \vec{OA}) \).
This is not right.
Positions relative to \( A \): \( A \) is at 0.
\( C \) is at \( k \).
\( B \) is at \( 4k \).
Let \( B \) divide \( AC \) in ratio \( m : n \).
\( B \) is \( \frac{mC + nA}{m+n} \).
Coordinates of \( B \) relative to \( A \): \( (4k) \).
Coordinates of \( C \) relative to \( A \): \( (k) \).
\[ 4k = \frac{m(k) + n(0)}{m+n} \implies 4k(m+n) = mk \] \[ 4(m+n) = m \implies 4m + 4n = m \implies 3m = -4n \implies \frac{m}{n} = -\frac{4}{3} \] So the ratio is \( -4 : 3 \).
Option \( 4 : -3 \) represents \( m = 4, n = -3 \) or \( m = -4, n = 3 \).
Option (3) is \( 4 : -3 \).