The Cartesian equation of a line is frac{x+3} | vedclass

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

Question

(A) Comparing the given Cartesian equation $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$ with the standard form $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\fr

Vedclass · Accepted Answer

$\vec{r}=(-3 \hat{i}+5 \hat{j}-6 \hat{k})+\lambda(2 \hat{i}+4 \hat{j}+2 \hat{k})$

Vedclass · Answer

(A) Comparing the given Cartesian equation $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$ with the standard form $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$,we get:$x_{1}=-3, y_{1}=5, z_{1}=-6$ and $a=2, b=4, c=2$.Thus,the line passes through the point $A(-3, 5, -6)$ and is parallel to the vector $\vec{b} = 2\hat{i} + 4\hat{j} + 2\hat{k}$.The position vector of the point is $\vec{a} = -3\hat{i} + 5\hat{j} - 6\hat{k}$.The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b}$.Substituting the values,we get $\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})$.

The Cartesian equation of a line is $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$. Find the vector equation for the line.

Similar Questions

The foot of the perpendicular drawn from the point $(1, 8, 4)$ on the line joining the points $(0, -11, 4)$ and $(2, -3, 1)$ is

The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is:

If the distances of the point $P(1, 2, a)$ from the line $L: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $L_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $L_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

The shortest distance between the lines $\vec{r} = (\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})$ and $\vec{r} = (-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}) + \mu(\hat{i} - \hat{k})$, where $\lambda, \mu \in R$, is:

The angle between two lines $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}$ and $\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module