The equation of the line passing through the | vedclass

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(D) The $XOZ$ plane is the plane where $y = 0$. The normal vector to the $XOZ$ plane is parallel to the $y$-axis,which is given by the vector $\vec{n}

Vedclass · Accepted Answer

$x = 2; \quad y = 3 + \lambda; \quad z = -4$

Vedclass · Answer

(D) The $XOZ$ plane is the plane where $y = 0$. The normal vector to the $XOZ$ plane is parallel to the $y$-axis,which is given by the vector $\vec{n} = (0, 1, 0)$.Since the required line is perpendicular to the $XOZ$ plane,it must be parallel to the normal vector $\vec{n} = (0, 1, 0)$.The line passes through the point $(2, 3, -4)$.Using the symmetric form of the line equation,$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$,where $(x_1, y_1, z_1) = (2, 3, -4)$ and the direction ratios are $(a, b, c) = (0, 1, 0)$.Thus,the equation is $\frac{x - 2}{0} = \frac{y - 3}{1} = \frac{z + 4}{0} = \lambda$.This implies $x - 2 = 0 \implies x = 2$,$z + 4 = 0 \implies z = -4$,and $y - 3 = \lambda \implies y = 3 + \lambda$.Therefore,the equation of the line is $x = 2, \quad y = 3 + \lambda, \quad z = -4$.

The equation of the line passing through the point $(2, 3, -4)$ and perpendicular to the $XOZ$ plane is

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