If the x-coordinate of a point P on the line | vedclass

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

Question

(B) Let the point $P$ divide the line segment $QR$ in the ratio $\lambda : 1$. The coordinates of $Q$ are $(2, 2, 1)$ and $R$ are $(5, 2, -2)$.Using t

Vedclass · Accepted Answer

$-2(z	ext{-coordinate of } P)$

Vedclass · Answer

(B) Let the point $P$ divide the line segment $QR$ in the ratio $\lambda : 1$. The coordinates of $Q$ are $(2, 2, 1)$ and $R$ are $(5, 2, -2)$.Using the section formula,the $x$-coordinate of $P$ is given by $x = \frac{5\lambda + 2}{\lambda + 1}$.Given $x = 4$,we have $4 = \frac{5\lambda + 2}{\lambda + 1} \Rightarrow 4\lambda + 4 = 5\lambda + 2 \Rightarrow \lambda = 2$.Now,find the $y$-coordinate of $P$: $y = \frac{2\lambda + 2}{\lambda + 1} = \frac{2(2) + 2}{2 + 1} = \frac{6}{3} = 2$.Next,find the $z$-coordinate of $P$: $z = \frac{-2\lambda + 1}{\lambda + 1} = \frac{-2(2) + 1}{2 + 1} = \frac{-3}{3} = -1$.Comparing the coordinates,we see that $y = 2$ and $z = -1$. Thus,$y = -2(z)$.Therefore,the $y$-coordinate of $P$ is $-2(z	ext{-coordinate of } P)$.

If the $x$-coordinate of a point $P$ on the line joining the points $Q(2, 2, 1)$ and $R(5, 2, -2)$ is $4$,then the $y$-coordinate of $P$ is:

Similar Questions

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4},$ $l \neq 0$ is $(3,5,7),$ then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to:

The equation of the line passing through $(1, 2, 3)$ and perpendicular to the lines $x-1 = \frac{y+2}{2} = \frac{z+4}{4}$ and $\frac{x-1}{2} = \frac{y-2}{2} = z+3$ is

Find the perpendicular distance from the point $P(4, -5, 3)$ to the line $\vec{r} = (5, -2, 6) + k(3, -4, 5)$,where $k \in \mathbb{R}$.

Find the shortest distance between the lines whose vector equations are $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$.

If $(a, b, c)$ are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$,then the point $P(a, 3b, 2c)$ lies on the plane:

Vedclass Test Series

Exam Paper Generator

Online Exam Module