Let the equation of the plane passing through | vedclass

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

Question

(C) The equation of the plane passing through the intersection of $x+2y+az-2=0$ and $x-y+z-3=0$ is given by $(x+2y+az-2) + \lambda(x-y+z-3) = 0$.This

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(C) The equation of the plane passing through the intersection of $x+2y+az-2=0$ and $x-y+z-3=0$ is given by $(x+2y+az-2) + \lambda(x-y+z-3) = 0$.This simplifies to $(1+\lambda)x + (2-\lambda)y + (a+\lambda)z - (2+3\lambda) = 0$.Comparing this with the given plane $5x-11y+bz = 6a-1$,we have:$\frac{1+\lambda}{5} = \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} = \frac{2+3\lambda}{6a-1}$.From $\frac{1+\lambda}{5} = \frac{2-\lambda}{-11}$,we get $-11-11\lambda = 10-5\lambda$,so $-6\lambda = 21$,which gives $\lambda = -\frac{7}{2}$.Substituting $\lambda = -\frac{7}{2}$ into the ratios:$\frac{1-3.5}{5} = -0.5$,so $\frac{2+3(-3.5)}{6a-1} = -0.5 \implies \frac{2-10.5}{6a-1} = -0.5 \implies \frac{-8.5}{6a-1} = -0.5 \implies 6a-1 = 17 \implies 6a = 18 \implies a = 3$.Now,$\frac{a+\lambda}{b} = -0.5 \implies \frac{3-3.5}{b} = -0.5 \implies \frac{-0.5}{b} = -0.5 \implies b = 1$.The plane is $5x-11y+z = 17$.The distance from $(a, -c, c) = (3, -c, c)$ to $5x-11y+z-17=0$ is $\frac{|5(3)-11(-c)+c-17|}{\sqrt{5^2+(-11)^2+1^2}} = \frac{|15+11c+c-17|}{\sqrt{25+121+1}} = \frac{|12c-2|}{\sqrt{147}}$.Given distance is $\frac{2}{\sqrt{a}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{49}}{\sqrt{147}} = \frac{14}{\sqrt{147}}$.So,$|12c-2| = 14$. This gives $12c-2 = 14 \implies 12c = 16$ (no integer solution) or $12c-2 = -14 \implies 12c = -12 \implies c = -1$.Thus,$\frac{a+b}{c} = \frac{3+1}{-1} = -4$.

Let the equation of the plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1$. For $c \in \mathbb{Z}$,if the distance of this plane from the point $(a, -c, c)$ is $\frac{2}{\sqrt{a}}$,then $\frac{a+b}{c}$ is equal to

Similar Questions

If the equation of the plane containing the line $x+2y+3z-4=0=2x+y-z+5$ and perpendicular to the plane $\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2\hat{j}+3\hat{k})$ is $ax+by+cz=4$,then $(a-b+c)$ is equal to

Let the plane containing the line of intersection of the planes $P_1: x+(\lambda+4)y+z=1$ and $P_2: 2x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane $P_2$ is (in $\sqrt{6}$)

The angle between the line $\frac{x - 1}{-2} = \frac{y - 2}{1} = \frac{z + 1}{2}$ and the plane $3x + 2y + 6z = 1$ is:

If the equation of the plane passing through the points $(2,1,2)$ and $(1,2,1)$ and perpendicular to the plane $2x - y + 2z = 1$ is $ax + by + cz + d = 0$, then $\frac{a+b}{c+d} = $

The point$(s)$ on the line $\vec r = \hat i + \hat j + \hat k + t(\hat i + 3\hat j - \hat k)$ at a distance of $3 \ units$ from the plane $\vec r \cdot (\hat i + 2\hat j + 2\hat k) + 2 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module