The point of intersection of the lines frac{x | vedclass

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(C) Let the first line be $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} = \lambda$. Any point on this line is $(3\lambda - 1, 5\lambda - 3, 7\l

Vedclass · Accepted Answer

$\left( \frac{1}{2}, - \frac{1}{2}, - \frac{3}{2} \right)$

Vedclass · Answer

(C) Let the first line be $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} = \lambda$. Any point on this line is $(3\lambda - 1, 5\lambda - 3, 7\lambda - 5)$. Let the second line be $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} = \mu$. Any point on this line is $(\mu + 2, 3\mu + 4, 5\mu + 6)$. For the lines to intersect,the coordinates must be equal: $3\lambda - 1 = \mu + 2 \implies 3\lambda - \mu = 3$ $5\lambda - 3 = 3\mu + 4 \implies 5\lambda - 3\mu = 7$ Solving these equations: Multiply the first by $3$: $9\lambda - 3\mu = 9$. Subtract the second from this: $(9\lambda - 3\mu) - (5\lambda - 3\mu) = 9 - 7 \implies 4\lambda = 2 \implies \lambda = \frac{1}{2}$. Substituting $\lambda = \frac{1}{2}$ into $3\lambda - \mu = 3$: $3(\frac{1}{2}) - \mu = 3 \implies \mu = \frac{3}{2} - 3 = -\frac{3}{2}$. Now,find the point using $\lambda = \frac{1}{2}$: $x = 3(\frac{1}{2}) - 1 = \frac{1}{2}$,$y = 5(\frac{1}{2}) - 3 = -\frac{1}{2}$,$z = 7(\frac{1}{2}) - 5 = -\frac{3}{2}$. The point of intersection is $(\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2})$.

The point of intersection of the lines $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$ is

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