The symmetric equation of the line formed by | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the direction ratios of the required line be $a, b, c$. Since the line lies in both planes,it is perpendicular to the normals of both planes. Thus,$3a + 2b + c = 0$ and $a + b - 2c = 0$. Using the cross product method,we have: $\frac{a}{(2)(-2) - (1)(1)} = \frac{b}{(1)(1) - (3)(-2)} = \frac{c}{(3)(1) - (2)(1)}$ $\frac{a}{-4-1} = \frac{b}{1+6} = \frac{c}{3-2}$ $\frac{a}{-5} = \frac{b}{7} = \frac{c}{1}$. To find a point on the line,we set $z = 0$ in the given plane equations: $3x + 2y = 5$ and $x + y = 3$. Solving these,we multiply the second by $2$: $2x + 2y = 6$. Subtracting this from the first: $(3x - 2x) = 5 - 6 \Rightarrow x = -1$. Substituting $x = -1$ into $x + y = 3$,we get $-1 + y = 3 \Rightarrow y = 4$. So,the point is $(-1, 4, 0)$. The symmetric equation is $\frac{x - (-1)}{-5} = \frac{y - 4}{7} = \frac{z - 0}{1}$,which is $\frac{x+1}{-5} = \frac{y-4}{7} = \frac{z-0}{1}$.

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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