Which of the following is correct to the line  5x+y+3z=0 and 3x−y+z+1=0

# Which of the following is correct to the line  $5x+y+3z=0$ and $3x-y+z+1=0$

1. A

Symmetrical form of the equation of line is$\frac{x}{2}=\frac{y-\frac{1}{8}}{-1}=\frac{z+\frac{5}{8}}{1}$

2. B

Symmetrical form of the equations of line is $\frac{x+\frac{1}{8}}{1}=\frac{y-\frac{5}{8}}{1}=\frac{z}{-2}$

3. C

Equation of the plane through $\left(2,-1,4\right)$ and perpendicular to the given line is $2x-y+z-7=0$

4. D

Equation of line through and perpendicular to the given line is$x+y-2z+5=0$

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### Solution:

The given planes are  $5x+y+3z=0$ and $3x-y+z+1=0$

Suppose that the direction ratios of line of intersection of the above two planes is $⟨a,b,c⟩$

Hence, $3a-b+c=0,5a+b+3c=0$$⇒\frac{a}{1}=\frac{b}{1}=\frac{c}{-2}$

To get a point on the line, substitute $z=0$  in the plane equations and the solve for the other variables

hence a point on the  line of intersection of planes is $P\left(-\frac{1}{8},\frac{5}{8},0\right)$

Therefore, the equation of the required line is $\frac{x+\frac{1}{8}}{1}=\frac{y-\frac{5}{8}}{1}=\frac{z}{-2}$

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