The line,that is coplanar to the line frac{x+ | vedclass

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

Question

(B) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar

Vedclass · Accepted Answer

$\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$

Vedclass · Answer

(B) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2\end{array}ight| = 0$.Given line $L_1: \frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$,point $P_1(-3, 1, 5)$ and direction vector $\vec{v_1} = (-3, 1, 5)$.For option $B$: Line $L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5}$,point $P_2(-1, 2, 5)$ and direction vector $\vec{v_2} = (-1, 2, 5)$.Vector $\vec{P_1P_2} = (-1 - (-3), 2 - 1, 5 - 5) = (2, 1, 0)$.Check coplanarity condition:$\left|\begin{array}{ccc}2 & 1 & 0 \ -3 & 1 & 5 \ -1 & 2 & 5\end{array}ight| = 2(5 - 10) - 1(-15 - (-5)) + 0 = 2(-5) - 1(-10) = -10 + 10 = 0$.Since the determinant is $0$,the lines are coplanar. Thus,option $B$ is correct.

The line,that is coplanar to the line $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$,is

Similar Questions

If the line joining $A(4,1,2)$ and $B(0, k, 1)$ is perpendicular to the line joining $C(-2,1,1)$ and $D(4,2,5)$,then the value of $k$ is equal to

The length of the perpendicular from the point $P(2, -1, 4)$ to the straight line $\frac{x + 3}{10} = \frac{y - 2}{-7} = \frac{z}{1}$ is

$A(2,3,4), B(4,5,7), C(2,-6,3), D(4,-4, k)$ are four points. If the line $\overline{AB}$ is parallel to $\overline{CD}$,then $k$ is equal to

The equation of the line of the shortest distance between the lines $\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$ and $\frac{x - 1}{0} = \frac{y + 1}{-2} = \frac{z}{1}$ is

The distance of the point $({x_1}, {y_1}, {z_1})$ from the line $\frac{{x - {x_2}}}{l} = \frac{{y - {y_2}}}{m} = \frac{{z - {z_2}}}{n}$,where $l, m, n$ are the direction cosines of the line,is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module