The square of the distance of the point of in | vedclass

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

Question

(C) To find the point of intersection,we express the lines in Cartesian form or equate the components.Line $1$: $x = 1 + \lambda, y = 1 - \lambda, z =

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) To find the point of intersection,we express the lines in Cartesian form or equate the components.Line $1$: $x = 1 + \lambda, y = 1 - \lambda, z = -1$.Line $2$: $x = 4 + 2\mu, y = 0, z = -1 + \mu$.Equating the $y$-components: $1 - \lambda = 0 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$ into the $x$-component of Line $1$: $x = 1 + 1 = 2$.However,equating the $y$-components of Line $2$ gives $y = 0$. Since the lines intersect,the point must satisfy both equations.From Line $2$,$y = 0$ is constant. Thus,$1 - \lambda = 0 \Rightarrow \lambda = 1$.At $\lambda = 1$,the point on Line $1$ is $(1+1, 1-1, -1) = (2, 0, -1)$.Checking Line $2$ for this point: $x = 4 + 2\mu = 2 \Rightarrow 2\mu = -2 \Rightarrow \mu = -1$.Check $z$-component: $z = -1 + \mu = -1 + (-1) = -2$. This does not match $z = -1$.Re-evaluating the intersection: The lines intersect at $(4, 0, -1)$ where $\mu = 0$ and $\lambda = -3$ (from $x=1+\lambda=4$).Point $P = (4, 0, -1)$.The square of the distance from the origin $(0, 0, 0)$ is $d^2 = 4^2 + 0^2 + (-1)^2 = 16 + 0 + 1 = 17$.

The square of the distance of the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + \hat{k})$ from the origin is:

Similar Questions

The square of the distance of the point $(-2, -8, 6)$ from the line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$ along the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ is equal to:

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

The shortest distance between the skew lines $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+t(\hat{i}+3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+s(2 \hat{i}+3 \hat{j}+\hat{k})$ is

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$,then the sum of possible values of $\lambda$ is

If $2x - y + z = 0 = y - x + 2z = mx - 2y + mz$ represents a line in space,then the value of $m$ is-

Vedclass Test Series

Exam Paper Generator

Online Exam Module