For the line frac{x + 1}{1} = frac{y - 2}{2} | vedclass

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(C) The given line is $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} = \lambda$. The general point is $(x, y, z) = (\lambda - 1, 2\lambda + 2, 3

Vedclass · Accepted Answer

It is perpendicular to the plane $x - 2y + z = 0$

Vedclass · Answer

(C) The given line is $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} = \lambda$. The general point is $(x, y, z) = (\lambda - 1, 2\lambda + 2, 3\lambda - 3)$.$(A)$ For $\lambda = -1$, the point is $(-2, 0, -6)$. The line $\frac{x + 2}{1} = \frac{y}{2} = \frac{z + 6}{3}$ passes through $(-2, 0, -6)$ with direction ratios $(1, 2, 3)$, which is the same line.$(B)$ Substitute the general point in the plane: $(\lambda - 1) - 2(2\lambda + 2) + (3\lambda - 3) + 8 = \lambda - 1 - 4\lambda - 4 + 3\lambda - 3 + 8 = 0$. Since it satisfies the equation, the line lies in the plane.$(C)$ The direction vector of the line is $\vec{v} = \hat{i} + 2\hat{j} + 3\hat{k}$ and the normal to the plane is $\vec{n} = \hat{i} - 2\hat{j} + \hat{k}$. Since $\vec{v} 
eq k\vec{n}$, the line is not perpendicular to the plane.$(D)$ For point $(0, 4, 0)$, $\frac{0+1}{1} = 1$, $\frac{4-2}{2} = 1$, $\frac{0+3}{3} = 1$. Since all ratios are equal, it passes through $(0, 4, 0)$.Thus, statement $(C)$ is incorrect.

For the line $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$, identify the incorrect statement among the following.

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