If the shortest distance between the lines fr | vedclass

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

Question

(C) The second line is given by the intersection of two planes: $x + y + z + 1 = 0$ and $2x - y + z + 3 = 0$.The equation of the family of planes pass

Vedclass · Accepted Answer

$\frac{32}{19}$

Vedclass · Answer

(C) The second line is given by the intersection of two planes: $x + y + z + 1 = 0$ and $2x - y + z + 3 = 0$.The equation of the family of planes passing through the line of intersection is $(x + y + z + 1) + \lambda(2x - y + z + 3) = 0$, which simplifies to $(1 + 2\lambda)x + (1 - \lambda)y + (1 + \lambda)z + (1 + 3\lambda) = 0$.The first line has direction vector $\vec{v_1} = (\alpha, -1, 1)$. The normal to the plane is $\vec{n} = (1 + 2\lambda, 1 - \lambda, 1 + \lambda)$.Since the line is parallel to the plane, $\vec{v_1} \cdot \vec{n} = 0$, so $\alpha(1 + 2\lambda) - (1 - \lambda) + (1 + \lambda) = 0$, which gives $\alpha(1 + 2\lambda) + 2\lambda = 0$, or $\alpha = -\frac{2\lambda}{1 + 2\lambda}$.The shortest distance between the line and the plane is the perpendicular distance from any point on the line (e.g., $(1, -1, 0)$) to the plane:$d = \frac{|(1 + 2\lambda)(1) + (1 - \lambda)(-1) + (1 + \lambda)(0) + (1 + 3\lambda)|}{\sqrt{(1 + 2\lambda)^2 + (1 - \lambda)^2 + (1 + \lambda)^2}} = \frac{1}{\sqrt{3}}$.Simplifying the numerator: $|1 + 2\lambda - 1 + \lambda + 1 + 3\lambda| = |6\lambda + 1|$.Simplifying the denominator: $\sqrt{1 + 4\lambda + 4\lambda^2 + 1 - 2\lambda + \lambda^2 + 1 + 2\lambda + \lambda^2} = \sqrt{6\lambda^2 + 4\lambda + 3}$.Squaring both sides: $\frac{(6\lambda + 1)^2}{6\lambda^2 + 4\lambda + 3} = \frac{1}{3} \Rightarrow 3(36\lambda^2 + 12\lambda + 1) = 6\lambda^2 + 4\lambda + 3$.$108\lambda^2 + 36\lambda + 3 = 6\lambda^2 + 4\lambda + 3 \Rightarrow 102\lambda^2 + 32\lambda = 0$.Thus, $\lambda = 0$ or $\lambda = -\frac{32}{102} = -\frac{16}{51}$.If $\lambda = 0$, $\alpha = 0$. If $\lambda = -\frac{16}{51}$, $\alpha = -\frac{2(-16/51)}{1 + 2(-16/51)} = \frac{32/51}{(51 - 32)/51} = \frac{32}{19}$.

If the shortest distance between the lines $\frac{x - 1}{\alpha} = \frac{y + 1}{-1} = \frac{z}{1}, (\alpha \ne -1)$ and $x + y + z + 1 = 0 = 2x - y + z + 3$ is $\frac{1}{\sqrt{3}}$, then a value of $\alpha$ is

Similar Questions

The equation of the plane containing the line $r = i + j + \lambda (2i + j + 4k)$ is

If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$ $(a > 0)$ makes an angle $30^{\circ}$ with the plane $y - z + 2 = 0$,then the direction cosines of the line are:

Let two vertices of triangle $ABC$ be $(2,4,6)$ and $(0,-2,-5)$,and its centroid be $(2,1,-1)$. If the image of the third vertex in the plane $x+2y+4z=11$ is $(\alpha, \beta, \gamma)$,then $\alpha \beta+\beta \gamma+\gamma \alpha$ is equal to

If the points $(1, 1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$, then the values of $\lambda$ are

What is the equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $3x + 4y - 5z = 6$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module