(A) Eliminating $t$ from the given equations,we get the equation of the path:$\frac{x}{2} = \frac{y}{-4} = \frac{z}{4}$ or $\frac{x}{1} = \frac{y}{-2}

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(A) Eliminating $t$ from the given equations,we get the equation of the path:$\frac{x}{2} = \frac{y}{-4} = \frac{z}{4}$ or $\frac{x}{1} = \frac{y}{-2} = \frac{z}{2}$.This represents a straight line passing through the origin.For $t = 10 \text{ s}$,the coordinates are:$x = 2(10) = 20 \text{ km}$,$y = -4(10) = -40 \text{ km}$,$z = 4(10) = 40 \text{ km}$.The distance from the origin $O(0, 0, 0)$ is given by:$d = \sqrt{x^2 + y^2 + z^2} = \sqrt{20^2 + (-40)^2 + 40^2}$$d = \sqrt{400 + 1600 + 1600} = \sqrt{3600} = 60 \text{ km}$.

Class 12 Mathematics THREE DIMENSIONAL GEOMETRY Line

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The equations of motion of a rocket are: $x = 2t, y = -4t, z = 4t$ where the time $t$ is given in seconds,and the coordinates of a moving point in kilometers. What is the path of the rocket? At what distance will the rocket be from the starting point $O(0, 0, 0)$ in $10$ seconds?

A
Straight line,$60 \text{ km}$
B
Straight line,$30 \text{ km}$
C
Parabola,$60 \text{ km}$
D
Ellipse,$60 \text{ km}$

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The vertices $B$ and $C$ of a $\Delta ABC$ lie on the line $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $BC = 5 \text{ units}$. If the vertex $A$ is $(1, -1, 2)$,then the area of this triangle (in $\text{sq. units}$) is:

DifficultJEE MAIN 2019

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The equation of the line that passes through the origin and is parallel to the $X$-axis is . . . . . . .

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Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.
Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

MediumAIEEE 2011

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Let $P(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z}{1}$ at a distance $4\sqrt{14}$ from the point $(1, -1, 0)$ and nearer to the origin. Then the shortest distance between the lines $\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}$ and $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$ is equal to

DifficultJEE MAIN 2026

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Find the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$.

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The equations of motion of a rocket are: $x = 2t, y = -4t, z = 4t$ where the time $t$ is given in seconds,and the coordinates of a moving point in kilometers. What is the path of the rocket? At what distance will the rocket be from the starting point $O(0, 0, 0)$ in $10$ seconds?

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The vertices $B$ and $C$ of a $\Delta ABC$ lie on the line $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $BC = 5 \text{ units}$. If the vertex $A$ is $(1, -1, 2)$,then the area of this triangle (in $\text{sq. units}$) is:

The equation of the line that passes through the origin and is parallel to the $X$-axis is . . . . . . .

Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

Find the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$.

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Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.
Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.