At what point does the line joining the point | vedclass

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(C) The equation of the line passing through $(x_1, y_1, z_1) = (2, -3, 1)$ and $(x_2, y_2, z_2) = (3, -4, -5)$ is given by $\frac{x-2}{3-2} = \frac{y

Vedclass · Accepted Answer

$(1, -2, 7)$

Vedclass · Answer

(C) The equation of the line passing through $(x_1, y_1, z_1) = (2, -3, 1)$ and $(x_2, y_2, z_2) = (3, -4, -5)$ is given by $\frac{x-2}{3-2} = \frac{y-(-3)}{-4-(-3)} = \frac{z-1}{-5-1} = k$.This simplifies to $\frac{x-2}{1} = \frac{y+3}{-1} = \frac{z-1}{-6} = k$.Any point on this line is given by $(k+2, -k-3, -6k+1)$.Since this point lies on the plane $2x + y + z = 7$,we substitute these coordinates into the plane equation:$2(k+2) + (-k-3) + (-6k+1) = 7$.$2k + 4 - k - 3 - 6k + 1 = 7$.$-5k + 2 = 7$.$-5k = 5 \implies k = -1$.Substituting $k = -1$ into the point coordinates:$x = -1 + 2 = 1$.$y = -(-1) - 3 = 1 - 3 = -2$.$z = -6(-1) + 1 = 6 + 1 = 7$.Thus,the point of intersection is $(1, -2, 7)$.

At what point does the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersect the plane $2x + y + z = 7$?

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