What is the point of intersection of the line | vedclass

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Question

(D) Let the line be $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3} = r$. Then,the coordinates of any point on the line are given by $x = r$,$y = 2r

Vedclass · Accepted Answer

$\left( -\frac{1}{11}, \frac{9}{11}, -\frac{25}{11} \right)$

Vedclass · Answer

(D) Let the line be $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3} = r$. Then,the coordinates of any point on the line are given by $x = r$,$y = 2r + 1$,and $z = 3r - 2$. Since this point lies on the plane $2x + 3y + z = 0$,we substitute these values into the plane equation: $2(r) + 3(2r + 1) + (3r - 2) = 0$. Expanding the equation: $2r + 6r + 3 + 3r - 2 = 0$. Combining like terms: $11r + 1 = 0$,which gives $r = -\frac{1}{11}$. Now,substitute $r = -\frac{1}{11}$ back into the expressions for $x, y, z$: $x = -\frac{1}{11}$,$y = 2(-\frac{1}{11}) + 1 = -\frac{2}{11} + \frac{11}{11} = \frac{9}{11}$,$z = 3(-\frac{1}{11}) - 2 = -\frac{3}{11} - \frac{22}{11} = -\frac{25}{11}$. Thus,the point of intersection is $\left( -\frac{1}{11}, \frac{9}{11}, -\frac{25}{11} \right)$.

What is the point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$?

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