The symmetric form of the line of intersection of two planes 3x−y=6,y+2z−1=0

The symmetric form of the line of intersection of two planes $3x-y=6,y+2z-1=0$

1. A

$\frac{x-\frac{7}{3}}{2}=\frac{y-1}{6}=\frac{z}{-3}$

2. B

$\frac{x-\frac{7}{3}}{2}=\frac{y+1}{6}=\frac{z}{3}$

3. C

$\frac{x+\frac{7}{3}}{2}=\frac{y-1}{6}=\frac{z}{-3}$

4. D

$\frac{x-2}{2}=\frac{y}{6}=\frac{z+0.5}{-3}$

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Solution:

The given planes are $3x-y=6,y+2z-1=0$

Plug in $z=0$ to get a point on the line.

$3x-y=6,y=1$

solve the above two equations, we get $x=\frac{7}{3},y=1$

A point on the line is $\left(\frac{7}{3},1,0\right)$

Vector along the line is cross product of two normal vectors to the planes

it gives ${n}_{1}×{n}_{2}=\left|\begin{array}{ccc}i& j& k\\ 3& -1& 0\\ 0& 1& 2\end{array}\right|=i\left(-2\right)-j\left(6\right)+k\left(3\right)$

Direction ratios of a line of intersection of two planes are proportional to $⟨2,6,-3⟩$

Therefore, the equation of the required line is $\overline{)\frac{x-\frac{7}{3}}{2}=\frac{y-1}{6}=\frac{z}{-3}}$

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