The Cartesian equation of a line is 2x - 3 = | vedclass

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

Question

(C) The given Cartesian equation is $2x - 3 = 3y + 1 = 5 - 6z$. To write this in standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}

Vedclass · Accepted Answer

$r = (7 \hat{i} - 5 \hat{j}) + \lambda(3 \hat{i} + 2 \hat{j} - \hat{k})$

Vedclass · Answer

(C) The given Cartesian equation is $2x - 3 = 3y + 1 = 5 - 6z$. To write this in standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$,we divide by the coefficients of $x, y, z$: $2(x - \frac{3}{2}) = 3(y + \frac{1}{3}) = -6(z - \frac{5}{6})$. Dividing by $6$,we get $\frac{x - 3/2}{3} = \frac{y + 1/3}{2} = \frac{z - 5/6}{-1}$. The direction vector of this line is $\vec{v} = 3 \hat{i} + 2 \hat{j} - \hat{k}$. The vector equation of a line passing through point $\vec{a} = 7 \hat{i} - 5 \hat{j} + 0 \hat{k}$ and parallel to $\vec{v}$ is given by $\vec{r} = \vec{a} + \lambda \vec{v}$. Thus,$\vec{r} = (7 \hat{i} - 5 \hat{j}) + \lambda(3 \hat{i} + 2 \hat{j} - \hat{k})$.

The Cartesian equation of a line is $2x - 3 = 3y + 1 = 5 - 6z$. The vector equation of the line passing through the point $(7, -5, 0)$ and parallel to the given line is

Similar Questions

Two lines whose direction cosines are given by $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ are perpendicular to each other if .........

If lines $\frac{x - 1}{3} = \frac{y - 2}{-1} = \frac{z - \lambda}{2}$ and $\frac{x + 1}{-2} = \frac{y}{3\lambda} = \frac{2z - 7}{1}$ are coplanar,then the sum of the value$(s)$ of $\lambda$ is:

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

If the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then $k$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module