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  • The line, that is coplanar to the line <img alt="\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx&plus;3%7D%7B-3%7D%3D%5Cfrac%7By-1%7

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

Option: 1

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


Option: 2

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


Option: 3

\frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}


Option: 4

\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}


Answers (1)

best_answer

Condition of co-planarity

\left|\begin{array}{lll} x_2-x_1 & a_1 & a_2 \\ y_2-y_1 & b_1 & b_2 \\ z_2-z_1 & c_1 & c_2 \end{array}\right|=0

Where a1, b1, c1 are direction cosine of 1^{\mathrm{st}} line and a2, b2, c2 are direction cosine of 2^{\mathrm{st}} line.

Now. Solving options

Point (-3, 1, 5) & point (-1, 2, 5)

$$ \begin{aligned} & \text { (1) }\left|\begin{array}{ccc} -3 & 1 & 5 \\ 1 & 2 & 5 \\ -2 & -1 & 0 \end{array}\right| \\ & =-3(5)-(10)+5(-1+4) \\ & =-15-10+15=-10 \\ & \end{aligned}\\ $$ (2) point $(-1,2,5)$\\\\ $$ \begin{aligned} & \left|\begin{array}{ccc} -3 & 1 & 5 \\ -1 & 2 & 5 \\ -2 & -1 & 0 \end{array}\right| \\ & =3(5)-(10)+5(1+4) \\ & -25+5=0 \end{aligned}\\\\ $$ (3) point (-1,2,5)

$$ \begin{aligned} & \left|\begin{array}{rcc} -3 & 1 & 5 \\ -1 & 2 & 4 \\ -2 & -1 & 0 \end{array}\right| \\ & -3(4)-(8)+5(1+4) \\ & -12-8+25=5 \end{aligned}\\\\

(4) \, \, point $(-1,2,5) $$

\begin{aligned} & \left|\begin{array}{rrr} -3 & 1 & 5 \\ -1 & 2 & 5 \\ 4 & 1 & 0 \end{array}\right| \\ & -3(-5)-(-20)+5(-1-8) \\ & 15+20-45=-10 \end{aligned}

Posted by

Deependra Verma

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