Let L be a line obtained from the intersectio | vedclass

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

Question

(D) The equations of the planes are $x+2y+z=6$ and $y+2z=4$. To find the line of intersection,we express $x$ and $y$ in terms of $z$. From $y+2z=4$,we

Vedclass · Accepted Answer

$102$

Vedclass · Answer

(D) The equations of the planes are $x+2y+z=6$ and $y+2z=4$. To find the line of intersection,we express $x$ and $y$ in terms of $z$. From $y+2z=4$,we get $y=4-2z$. Substituting this into the first plane equation: $x+2(4-2z)+z=6 \Rightarrow x+8-4z+z=6 \Rightarrow x-3z=-2 \Rightarrow x=3z-2$. Thus,the line $L$ can be written in symmetric form as $\frac{x+2}{3} = \frac{y-4}{-2} = z = \lambda$. Any point $P$ on the line $L$ is given by $(3\lambda-2, -2\lambda+4, \lambda)$. The direction vector of the line is $\vec{v} = 3\hat{i}-2\hat{j}+\hat{k}$. Let $A = (3,2,1)$. The vector $\vec{AP} = (3\lambda-2-3, -2\lambda+4-2, \lambda-1) = (3\lambda-5, -2\lambda+2, \lambda-1)$. Since $\vec{AP} \perp \vec{v}$,their dot product is zero: $(3\lambda-5)(3) + (-2\lambda+2)(-2) + (\lambda-1)(1) = 0$. $9\lambda - 15 + 4\lambda - 4 + \lambda - 1 = 0 \Rightarrow 14\lambda - 20 = 0 \Rightarrow \lambda = \frac{10}{7}$. Substituting $\lambda$ back,$P = (3(\frac{10}{7})-2, -2(\frac{10}{7})+4, \frac{10}{7}) = (\frac{16}{7}, \frac{8}{7}, \frac{10}{7})$. Thus,$\alpha+\beta+\gamma = \frac{16+8+10}{7} = \frac{34}{7}$. The value of $21(\alpha+\beta+\gamma) = 21 	imes \frac{34}{7} = 3 	imes 34 = 102$.

Let $L$ be a line obtained from the intersection of two planes $x+2y+z=6$ and $y+2z=4$. If point $P(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $(3,2,1)$ on $L$,then the value of $21(\alpha+\beta+\gamma)$ equals ...... .

Similar Questions

The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to

The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

Find the distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{-6}$.

The equation of the plane passing through the intersection of the planes $x + 2y + 3z + 4 = 0$ and $4x + 3y + 2z + 1 = 0$ and the origin is

$A$ perpendicular is drawn from a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x - y + z = 3$. Then the coordinates of $Q$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module