The vector equation of the line whose Cartesi | vedclass

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

Question

(B) Given Cartesian equations are $y=2$ and $4x-3z+5=0$. We can rewrite the second equation as $4x = 3z - 5$,which implies $4x = 3(z - \frac{5}{3})$.

Vedclass · Accepted Answer

$\overline{r}=(2\hat{j}+\frac{5}{3}\hat{k})+\lambda(3\hat{i}+4\hat{k})$

Vedclass · Answer

(B) Given Cartesian equations are $y=2$ and $4x-3z+5=0$. We can rewrite the second equation as $4x = 3z - 5$,which implies $4x = 3(z - \frac{5}{3})$. Dividing by $12$,we get $\frac{4x}{12} = \frac{3(z - \frac{5}{3})}{12}$,which simplifies to $\frac{x}{3} = \frac{z - \frac{5}{3}}{4}$. Since $y=2$ is constant,the line can be represented as $\frac{x}{3} = \frac{y-2}{0} = \frac{z - \frac{5}{3}}{4}$. This line passes through the point $(0, 2, \frac{5}{3})$,so its position vector is $\overline{a} = 0\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k}$. The direction ratios of the line are $(3, 0, 4)$,so the direction vector is $\overline{b} = 3\hat{i} + 0\hat{j} + 4\hat{k}$. The vector equation of a line is given by $\overline{r} = \overline{a} + \lambda\overline{b}$. Substituting the values,we get $\overline{r} = (2\hat{j} + \frac{5}{3}\hat{k}) + \lambda(3\hat{i} + 4\hat{k})$.

The vector equation of the line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

Similar Questions

If $A(1,2,3), B(2,3,-1), C(3,-1,-2)$ are the vertices of a triangle $ABC$,then the direction ratios of the internal angle bisector of $\angle A$ are:

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

If a point $P$ on the line segment joining the points $(3, 5, -1)$ and $(6, 3, -2)$ has its $y$-coordinate $2$,then its $z$-coordinate is

The vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$ is . . . . . . .

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is

Vedclass Test Series

Exam Paper Generator

Online Exam Module