On which of the following lines does the poin | vedclass

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Question

(A) Let the line be $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=\lambda$. Any point on this line is given by $(2\lambda+4, 2\lambda+5, \lambda+3)$. Sin

Vedclass · Accepted Answer

$\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$

Vedclass · Answer

(A) Let the line be $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=\lambda$. Any point on this line is given by $(2\lambda+4, 2\lambda+5, \lambda+3)$. Since this point lies on the plane $x+y+z=2$,we substitute the coordinates into the plane equation: $(2\lambda+4) + (2\lambda+5) + (\lambda+3) = 2$. $5\lambda + 12 = 2$. $5\lambda = -10$,so $\lambda = -2$. Substituting $\lambda = -2$ back into the point coordinates: $x = 2(-2) + 4 = 0$,$y = 2(-2) + 5 = 1$,$z = (-2) + 3 = 1$. The point of intersection is $(0, 1, 1)$. Now,check which option satisfies the point $(0, 1, 1)$: For option $(A)$: $\frac{0-1}{1} = -1$,$\frac{1-3}{2} = -1$,$\frac{1+4}{-5} = -1$. Since all ratios are equal to $-1$,the point $(0, 1, 1)$ lies on the line given in option $(A)$.

On which of the following lines does the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ lie?

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