The plane containing the line frac{x - 1}{1} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Since the plane contains the li

Vedclass · Accepted Answer

$(1, 0, 5)$

Vedclass · Answer

(B) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Since the plane contains the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$,it passes through $(1, 2, 3)$ and its normal vector $\vec{n} = (a, b, c)$ is perpendicular to the line's direction vector $\vec{v_1} = (1, 2, 3)$. Thus,$a(1) + b(2) + c(3) = 0 \implies a + 2b + 3c = 0$ $(i)$. Since the plane is parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$,its normal vector is also perpendicular to the line's direction vector $\vec{v_2} = (1, 1, 4)$. Thus,$a(1) + b(1) + c(4) = 0 \implies a + b + 4c = 0$ $(ii)$. Solving $(i)$ and $(ii)$ using cross product: $\frac{a}{8-3} = \frac{b}{3-4} = \frac{c}{1-2} = k$. So,$a = 5k, b = -k, c = -k$. Substituting into the plane equation: $5(x - 1) - 1(y - 2) - 1(z - 3) = 0$. $5x - 5 - y + 2 - z + 3 = 0 \implies 5x - y - z = 0$. Checking the options,for $(1, 0, 5)$: $5(1) - 0 - 5 = 0$. Thus,the plane passes through $(1, 0, 5)$.

The plane containing the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$ passes through the point

Similar Questions

For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to:

The distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ is

The distance from the point $(-1, -5, -10)$ to the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5$ is:

The equation of the plane passing through the origin and containing the line $\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{5}$ is:

The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies in the plane $2x-4y+z=7$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module