The symmetric equation of the line formed by | vedclass

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Question

(C) Let $a, b, c$ be the direction ratios of the required line.Since the line lies in both planes,its direction vector is perpendicular to the normals

Vedclass · Accepted Answer

$\frac{x + 1}{-5} = \frac{y - 4}{7} = \frac{z - 0}{1}$

Vedclass · Answer

(C) Let $a, b, c$ be the direction ratios of the required line.Since the line lies in both planes,its direction vector is perpendicular to the normals of both planes,$(3, 2, 1)$ and $(1, 1, -2)$.Thus,$3a + 2b + c = 0$ and $a + b - 2c = 0$.Using cross product,$\frac{a}{(-4 - 1)} = \frac{b}{(1 + 6)} = \frac{c}{(3 - 2)}$,which gives $\frac{a}{-5} = \frac{b}{7} = \frac{c}{1}$.To find a point on the line,set $z = 0$ in the given equations: $3x + 2y = 5$ and $x + y = 3$.Solving these,we get $x = -1$ and $y = 4$.Thus,the point is $(-1, 4, 0)$.The symmetric equation is $\frac{x - (-1)}{-5} = \frac{y - 4}{7} = \frac{z - 0}{1}$,which simplifies to $\frac{x + 1}{-5} = \frac{y - 4}{7} = \frac{z - 0}{1}$.

The symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

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