The line of intersection of the planes x + 2y | vedclass

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(B) The given planes are $x + 2y = 0$ and $y - 3z + 3 = 0$. From the first equation,$y = -\frac{x}{2}$. From the second equation,$y = 3z - 3$,which im

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given planes are $x + 2y = 0$ and $y - 3z + 3 = 0$. From the first equation,$y = -\frac{x}{2}$. From the second equation,$y = 3z - 3$,which implies $z = \frac{y+3}{3}$. Equating the expressions for $y$,we have $y = -\frac{x}{2} = 3(z-1)$. This can be rewritten as $\frac{x}{-2} = \frac{y}{1} = \frac{z-1}{1/3}$. Multiplying the denominators by $3$,we get $\frac{x}{-6} = \frac{y}{3} = \frac{z-1}{1}$. Comparing this with the given options,the line is represented by $\frac{x}{-6} = \frac{y}{3} = \frac{z-1}{1}$. Note: The provided option $A$ is $\frac{x}{-6} = \frac{y}{3} = \frac{z}{1}$,which is equivalent to the direction vector $( -6, 3, 1 )$. Since the line must pass through a point satisfying both planes (e.g.,$( -6, 3, 2 )$),the correct symmetric form is $\frac{x+6}{-6} = \frac{y-3}{3} = \frac{z-2}{1}$.

The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is

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