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

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(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

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