If the equation of the plane containing the l | vedclass

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

Question

(C) The line is given by the intersection of planes $P_1: x+2y+3z-4=0$ and $P_2: 2x+y-z+5=0$. The direction vector $\vec{v}$ of the line is $\vec{n}_1

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(C) The line is given by the intersection of planes $P_1: x+2y+3z-4=0$ and $P_2: 2x+y-z+5=0$. The direction vector $\vec{v}$ of the line is $\vec{n}_1 	imes \vec{n}_2$,where $\vec{n}_1 = \hat{i}+2\hat{j}+3\hat{k}$ and $\vec{n}_2 = 2\hat{i}+\hat{j}-\hat{k}$.$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 2 & 1 & -1 \end{vmatrix} = \hat{i}(-2-3) - \hat{j}(-1-6) + \hat{k}(1-4) = -5\hat{i}+7\hat{j}-3\hat{k}$.The second plane is given in parametric form $\vec{r} = \vec{a} + \lambda\vec{u} + \mu\vec{w}$,where $\vec{u} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{w} = \hat{i}-2\hat{j}+3\hat{k}$. The normal vector $\vec{n}_3$ to this plane is $\vec{u} 	imes \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & -2 & 3 \end{vmatrix} = \hat{i}(3+2) - \hat{j}(3-1) + \hat{k}(-2-1) = 5\hat{i}-2\hat{j}-3\hat{k}$.The required plane contains the line (direction $\vec{v}$) and is perpendicular to the second plane (normal $\vec{n}_3$). Thus,the normal $\vec{N}$ of the required plane is $\vec{v} 	imes \vec{n}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -5 & 7 & -3 \ 5 & -2 & -3 \end{vmatrix} = \hat{i}(-21-6) - \hat{j}(15+15) + \hat{k}(10-35) = -27\hat{i}-30\hat{j}-25\hat{k}$.Taking the normal as $27\hat{i}+30\hat{j}+25\hat{k}$,the plane equation is $27x+30y+25z=d$. $A$ point on the line (set $z=0$) is $x+2y=4$ and $2x+y=-5$,giving $x=-14/3, y=13/3$. Substituting into the plane equation: $27(-14/3) + 30(13/3) + 25(0) = -126 + 130 = 4$. So $d=4$.Thus,$a=27, b=30, c=25$. Then $a-b+c = 27-30+25 = 22$.

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

Similar Questions

Let the foot of the perpendicular from the point $P (3, -2, -9)$ on the plane passing through the points $A (-1, -2, -3)$,$B (9, 3, 4)$,and $C (9, -2, 1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is:

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

$A$ 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 $Q$. Then the length of the line segment $PQ$ equals

In what ratio does the $xy$-plane divide the line segment joining the points $(1, 2, 3)$ and $(4, 2, 1)$?

Angle between the lines of intersection of the planes $x-y=0, 2x+y+z=0$ and $2x-z=0, x+y-3z=0$ is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module