The acute angle between the lines x=-y, z=0 a | vedclass

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

Question

(C) The given lines are $x = -y, z = 0$ and $x = 0, z = 0$. We can write these lines in symmetric form: Line $1$: $\frac{x}{1} = \frac{y}{-1} = \frac{

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) The given lines are $x = -y, z = 0$ and $x = 0, z = 0$. We can write these lines in symmetric form: Line $1$: $\frac{x}{1} = \frac{y}{-1} = \frac{z}{0}$,so the direction vector is $\vec{v_1} = (1, -1, 0)$. Line $2$: $\frac{x}{0} = \frac{y}{1} = \frac{z}{0}$,so the direction vector is $\vec{v_2} = (0, 1, 0)$. The angle $	heta$ between the lines is given by $\cos 	heta = \left| \frac{\vec{v_1} \cdot \vec{v_2}}{|\vec{v_1}| |\vec{v_2}|} ight|$. Calculating the dot product: $\vec{v_1} \cdot \vec{v_2} = (1)(0) + (-1)(1) + (0)(0) = -1$. Calculating the magnitudes: $|\vec{v_1}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$ and $|\vec{v_2}| = \sqrt{0^2 + 1^2 + 0^2} = 1$. Thus,$\cos 	heta = \left| \frac{-1}{\sqrt{2} 	imes 1} ight| = \frac{1}{\sqrt{2}}$. Therefore,$	heta = \cos^{-1}\left(\frac{1}{\sqrt{2}}ight) = \frac{\pi}{4}$.

The acute angle between the lines $x=-y, z=0$ and $x=0, z=0$ is

Similar Questions

Let $(\alpha, \beta, \gamma)$ be the image of the point $A(8, 5, 7)$ in the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$. Then $\alpha + \beta + \gamma$ is equal to

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

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

Prove that the lines $x=p y+q, z=r y+s$ and $x=p^{\prime} y+q^{\prime}, z=r^{\prime} y+s^{\prime}$ are perpendicular if $p p^{\prime}+r r^{\prime}+1=0$.

The distance of the point $P(1, 2, 1)$ from the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module