The Cartesian equation of a line is frac{x-5} | vedclass

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

Question

(B) The given Cartesian equation is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2}$.First,rewrite the equation in the standard form $\frac{x-x_1}{a} =

Vedclass · Accepted Answer

$\vec{r} = 5\hat{i} - 4\hat{j} + 6\hat{k} + \lambda(3\hat{i} + 7\hat{j} - 2\hat{k})$

Vedclass · Answer

(B) The given Cartesian equation is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2}$.First,rewrite the equation in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the third term,$\frac{6-z}{2} = \frac{-(z-6)}{2} = \frac{z-6}{-2}$.So,the equation becomes $\frac{x-5}{3} = \frac{y-(-4)}{7} = \frac{z-6}{-2}$.Comparing this with the standard form,the point $(x_1, y_1, z_1)$ is $(5, -4, 6)$ and the direction ratios $(a, b, c)$ are $(3, 7, -2)$.The position vector of the point is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$ and the direction vector is $\vec{b} = 3\hat{i} + 7\hat{j} - 2\hat{k}$.The vector equation of a line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$.Substituting the values,we get $\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} - 2\hat{k})$.Thus,the correct option is $B$.

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{6-z}{2}$. Find the vector equation of the line.

Similar Questions

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 each other,then the value of $k$ is:

The values of $p$ and $q$ such that the line joining the points $(7, p, 2)$ and $(q, -2, 5)$ is parallel to the line joining the points $(2, -3, 5)$ and $(-6, -15, 11)$ are:

Let the line $L$ pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $P(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is:

The equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to each of the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

$A$ triangle $ABC$ is formed by vertices $A(1, -1, 0)$,$B(3, 5, 3)$,and $C(-11, -5, 6)$. The equation of the internal angle bisector of $\angle A$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module