The lines frac{x - 1}{-1} = frac{y + 2}{1} = | vedclass

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

Question

(D) Step $1$: Identify the direction vectors of the two lines. The direction vector of the first line is $\vec{b_1} = -1\hat{i} + 1\hat{j} - 2\hat{k}$

Vedclass · Accepted Answer

skew lines

Vedclass · Answer

(D) Step $1$: Identify the direction vectors of the two lines. The direction vector of the first line is $\vec{b_1} = -1\hat{i} + 1\hat{j} - 2\hat{k}$ and the direction vector of the second line is $\vec{b_2} = 1\hat{i} + 1\hat{j} - 2\hat{k}$.Step $2$: Check for perpendicularity by calculating the dot product $\vec{b_1} \cdot \vec{b_2} = (-1)(1) + (1)(1) + (-2)(-2) = -1 + 1 + 4 = 4$. Since the dot product is not $0$, the lines are not perpendicular.Step $3$: Check for intersection. Both lines pass through the point $(1, -2, z)$. For the first line, at $x=1, y=-2$, we have $z=3$. For the second line, at $x=1, y=-2$, we have $z=-1$. Since the lines pass through the same point $(1, -2, z)$ only if $z$ values match, and they do not, we check if they intersect at any other point. Setting the parametric equations equal: $1-t_1 = 1+t_2 \implies t_1+t_2=0$ and $-2+t_1 = -2+t_2 \implies t_1=t_2$. Thus $t_1=t_2=0$. At $t_1=0, t_2=0$, the points are $(1, -2, 3)$ and $(1, -2, -1)$. These are distinct points. Since the lines share a common point $(1, -2)$ in the $xy$-plane but have different $z$-coordinates, they are skew lines.

The lines $\frac{x - 1}{-1} = \frac{y + 2}{1} = \frac{z - 3}{-2}$ and $\frac{x - 1}{1} = \frac{y + 2}{1} = \frac{z + 1}{-2}$ are

Similar Questions

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-2}{1}=\frac{y+m}{2}=\frac{z-2}{1}$ intersect each other,then the value of $m$ is

The point of intersection of the lines $l_1: r(t) = (i - 6j + 2k) + t(i + 2j + k)$ and $l_2: R(u) = (4j + k) + u(2i + j + 2k)$ is

Distance between the parallel lines $\frac{x}{3}=\frac{y-1}{-2}=\frac{z}{1}$ and $\frac{x+4}{3}=\frac{y-3}{-2}=\frac{z+2}{1}$ is

If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$,then the sum of all possible values of $\lambda$ is :

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is $\qquad$

Vedclass Test Series

Exam Paper Generator

Online Exam Module