If the point (a, 8, -2) divides the line segm | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that the point $(a, 8, -2)$ divides the line segment joining $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m: n$. By the section formula,the co

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given that the point $(a, 8, -2)$ divides the line segment joining $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m: n$. By the section formula,the coordinates are given by: $x = \frac{5m + n}{m + n}$,$y = \frac{2m + 4n}{m + n}$,$z = \frac{10m + 6n}{m + n}$. Equating the $y$-coordinate: $8 = \frac{2m + 4n}{m + n} \implies 8m + 8n = 2m + 4n \implies 6m = -4n \implies \frac{m}{n} = -\frac{2}{3}$. Equating the $z$-coordinate: $-2 = \frac{10m + 6n}{m + n} \implies -2m - 2n = 10m + 6n \implies -12m = 8n \implies \frac{m}{n} = -\frac{8}{12} = -\frac{2}{3}$. Since both equations yield the same ratio,we find $a$ using the $x$-coordinate: $a = \frac{5m + n}{m + n} = \frac{5(\frac{m}{n}) + 1}{(\frac{m}{n}) + 1} = \frac{5(-\frac{2}{3}) + 1}{(-\frac{2}{3}) + 1} = \frac{-\frac{10}{3} + 1}{\frac{1}{3}} = \frac{-\frac{7}{3}}{\frac{1}{3}} = -7$. Finally,calculating the expression: $\frac{2m}{n} - \frac{a}{3} = 2(-\frac{2}{3}) - (\frac{-7}{3}) = -\frac{4}{3} + \frac{7}{3} = \frac{3}{3} = 1$.

If the point $(a, 8, -2)$ divides the line segment joining the points $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m: n$,then $\frac{2m}{n} - \frac{a}{3} =$

Similar Questions

Given that $P(3, 2, -4)$,$Q(5, 4, -6)$,and $R(9, 8, -10)$ are collinear. Find the ratio in which $Q$ divides $PR$.

The ratio in which the $yz$-plane divides the line segment joining the points $(-3, 4, -2)$ and $(2, 1, 3)$ is:

The ratio in which the $yz$-plane divides the line segment joining $(-3, 4, -2)$ and $(2, 1, 3)$ is

$A$ point $P(5, -1)$ divides the line segment joining points $A(11, -3)$ and $B(x, y)$ in the ratio $2:3$. Find the coordinates of point $B$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module