If the position vectors of two points P and Q | vedclass

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

Question

(B) Let the ratio in which the $YOZ$ plane divides the line segment $PQ$ be $k:1$. The coordinates of point $P$ are $(9, -1, 5)$ and point $Q$ are $(1

Vedclass · Accepted Answer

$-9:1$

Vedclass · Answer

(B) Let the ratio in which the $YOZ$ plane divides the line segment $PQ$ be $k:1$. The coordinates of point $P$ are $(9, -1, 5)$ and point $Q$ are $(1, 3, 5)$. The $YOZ$ plane is defined by the equation $x = 0$. Using the section formula,the $x$-coordinate of the point $R$ dividing $PQ$ in the ratio $k:1$ is given by: $x = \frac{k(x_2) + 1(x_1)}{k+1} = 0$ Substituting the values: $\frac{k(1) + 1(9)}{k+1} = 0$ $k + 9 = 0$ $k = -9$ Thus,the ratio $PR : RQ$ is $-9 : 1$.

If the position vectors of two points $P$ and $Q$ are respectively $9\hat{i} - \hat{j} + 5\hat{k}$ and $\hat{i} + 3\hat{j} + 5\hat{k}$,and the line segment $PQ$ intersects the $YOZ$ plane at a point $R$,then the ratio $PR : RQ$ is equal to

Similar Questions

Let the point,on the line passing through the points $P(1, -2, 3)$ and $Q(5, -4, 7)$,farther from the origin and at a distance of $9$ units from the point $P$,be $(\alpha, \beta, \gamma)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

If $\theta$ is the angle between the lines $\frac{x - 1}{2} = \frac{2y + 3}{4}; z = -2$ and $x = 1; \frac{y - 1}{2} = \frac{z + 1}{2}$, then

The distance between the parallel lines $\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z-3}{1}$ and $\frac{x}{2}=\frac{y}{-2}=\frac{z}{1}$ is

If $\frac{x - 1}{l} = \frac{y - 2}{m} = \frac{z + 1}{n}$ is the equation of the line passing through the points $(1, 2, -1)$ and $(-1, 0, 1)$,then the direction ratios $(l, m, n)$ are:

Direction ratios of the line represented by the equations $x = ay + b$ and $z = cy + d$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module