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 angle between the lines $\frac{x-2}{2} = \frac{y-3}{-2} = \frac{z-5}{1}$ and $\frac{x-2}{1} = \frac{y-3}{2} = \frac{z-5}{2}$ is $ . . . . . . $. (in $^{\circ}$)

If the image of the point $P(1, 0, 3)$ in the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$ is $Q(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma$ is equal to

If $A(1, 0, 2)$,$B(2, 1, 0)$,$C(2, -5, 3)$,and $D(0, 3, 2)$ are four points and the point of intersection of the lines $AB$ and $CD$ is $P(a, b, c)$,then $a + b + c =$

If the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then $k$ has the value

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then what is the value of $k$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module