Let the line frac{x-3}{7}=frac{y-2}{-1}=frac{ | vedclass

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

Question

(C) The plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$,which passes through $(4, -1, 0)$ and has direction vector $\vec{v_1} = \la

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$,which passes through $(4, -1, 0)$ and has direction vector $\vec{v_1} = \langle 1, -2, 1 \rangle$. The plane also contains the line defined by $4ax-y+5z-7a=0$ and $2x-5y-z-3=0$. The family of planes passing through the intersection of these two planes is $(4ax-y+5z-7a) + \lambda(2x-5y-z-3) = 0$. Since the point $(4, -1, 0)$ lies on the plane,we have $(16a+1-7a) + \lambda(8+5-3) = 0$,which simplifies to $9a + 10\lambda + 1 = 0$ (Equation $1$). The normal vector of the plane is $\vec{n} = \langle 4a+2\lambda, -1-5\lambda, 5-\lambda \rangle$. Since the line $\vec{v_1}$ lies in the plane,$\vec{n} \cdot \vec{v_1} = 0$,so $(4a+2\lambda) - 2(-1-5\lambda) + (5-\lambda) = 0$,which simplifies to $4a + 11\lambda + 7 = 0$ (Equation $2$). Solving Equations $1$ and $2$ gives $a=1$ and $\lambda=-1$. Substituting these into the plane equation: $(4x-y+5z-7) - (2x-5y-z-3) = 0$,which simplifies to $2x+4y+6z-4=0$ or $x+2y+3z-2=0$. The line is $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}=t$,so any point on it is $(7t+3, -t+2, -4t+3)$. Substituting this into the plane equation: $(7t+3) + 2(-t+2) + 3(-4t+3) - 2 = 0$,which gives $7t+3-2t+4-12t+9-2=0$,so $-7t+14=0$,implying $t=2$. The point $P$ is $(17, 0, -5)$. Thus,$\alpha+\beta+\gamma = 17+0-5 = 12$.

Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3, a \in R$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals...

Similar Questions

If the lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar,then the equation of the plane containing them is .........

Point $P$ is the intersection of the line joining points $Q(2, 3, 5)$ and $R(1, -1, 4)$ with the plane $5x - 4y - z = 1$. If $S$ is the foot of the perpendicular drawn from point $T(2, 1, 4)$ to the line $QR$,find the length of the line segment $PS$.

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ is equal to:

The image of the point with position vector $\vec{i} + 3\vec{k}$ in the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$ is

The ratio in which the line joining $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module