Find the equation of the line passing through | vedclass

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(B) The equation of the plane is given by $2x + 3y - z - 5 = 0$. The normal vector to this plane is $\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}$. Since t

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The equation of the plane is given by $2x + 3y - z - 5 = 0$. The normal vector to this plane is $\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}$. Since the line is perpendicular to the plane,the direction vector of the line is parallel to the normal vector of the plane. Thus,the direction ratios of the line are $(a, b, c) = (2, 3, -1)$. The line passes through the point $(x_1, y_1, z_1) = (1, 1, 1)$. The symmetric form of the equation of a line is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. Substituting the values,we get $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{-1}$. Therefore,the correct option is $B$.

Find the equation of the line passing through $(1, 1, 1)$ and perpendicular to the plane $2x + 3y - z - 5 = 0$.

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