On a line with direction cosines l, m, n,A(x_ | vedclass

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(A) Let the point $B$ divide the line segment $AC$ in the ratio $\lambda : 1$. Using the section formula,the coordinates of $B$ are given by: $B = \le

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$4: -3$

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(A) Let the point $B$ divide the line segment $AC$ in the ratio $\lambda : 1$. Using the section formula,the coordinates of $B$ are given by: $B = \left( \frac{\lambda(x_1 + kl) + 1(x_1)}{\lambda + 1}, \frac{\lambda(y_1 + km) + 1(y_1)}{\lambda + 1}, \frac{\lambda(z_1 + kn) + 1(z_1)}{\lambda + 1} ight)$. Comparing this with the given coordinates of $B = (x_1 + 4kl, y_1 + 4km, z_1 + 4kn)$,we equate the $x$-coordinates: $x_1 + 4kl = \frac{\lambda x_1 + \lambda kl + x_1}{\lambda + 1} = \frac{x_1(\lambda + 1) + \lambda kl}{\lambda + 1} = x_1 + \frac{\lambda kl}{\lambda + 1}$. Subtracting $x_1$ from both sides: $4kl = \frac{\lambda kl}{\lambda + 1}$. Since $k > 0$ and $l$ is a direction cosine (assuming $l 
eq 0$),we can divide by $kl$: $4 = \frac{\lambda}{\lambda + 1}$. $4(\lambda + 1) = \lambda \implies 4\lambda + 4 = \lambda \implies 3\lambda = -4 \implies \lambda = -\frac{4}{3}$. Thus,the ratio is $-4:3$ or $4:-3$. Therefore,the point $B$ divides the segment $AC$ externally in the ratio $4:3$.

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 $C = (x_1 + kl, y_1 + km, z_1 + kn)$ where $k > 0$,then the ratio in which the point $B$ divides the line segment joining $A$ and $C$ is:

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