Three Dimensional Geometry · Mathematics · COMEDK

The equation of the perpendicular drawn from the point $A(6,1,3)$ to the line $\frac{x-1}{2}=\frac{2-y}{-1}=\frac{z-3}{2}$ is $\frac{x-6}{\boldsymbol{a}}=\frac{y-1}{\boldsymbol{b}}=\frac{z-3}{\boldsymbol{c}}$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are the possible integers such that $\boldsymbol{a}<0$, then the value of $\boldsymbol{a}-\boldsymbol{b}+\mathbf{5} \boldsymbol{c}$ is:

$$ \begin{aligned} &\text { Consider two skew lines in 3D space. }\\ &M_1: \frac{x-1}{1}=\frac{2-y}{1}=\frac{z-5}{1} \text { and } M_2: \frac{x+3}{1}=\frac{y-7}{2}=\frac{z+4}{1} \end{aligned} $$

Let $L_1$ be the line of shortest distance (common perpendicular) between $M_1$ and $M_2$

If $L_2$ is a line parallel to the vector $\vec{b}=\hat{\jmath}+\hat{k}$,

Then the acute angle $\boldsymbol{\theta}$ between the lines $L_1$ and $L_2$ is:

Consider the lines $L_1$ and $L_2$ given by the following vector equations:

$$ L_1: \vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}+\boldsymbol{t} \hat{j}) \quad L_2: \vec{r}=(4 \hat{i}+\boldsymbol{a} \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{k}) $$

If $\boldsymbol{a}=-2$ and the lines intersect, then the value of ' $\mathbf{t}$ ' is:

The angle between the two lines whose direction cosines satisfy the relations $\boldsymbol{l}+\boldsymbol{m}+\boldsymbol{n}=\mathbf{0}$ and $\boldsymbol{l}^{\mathbf{2}}=\boldsymbol{m}^{\mathbf{2}}+\boldsymbol{n}^{\mathbf{2}}$ is

Let L be the foot of the perpendicular drawn from the point $P(5,3 k-7,-4)$ to the YZ - plane. If the distance of point L from the origin is $\sqrt{41}$ units, then the possible value of ' $\boldsymbol{k}$ ' is:

Let $\mathbf{P}$ be a point on the line $L_1: \frac{x-2}{2}=y+1=\frac{z-1}{2}$ such that its distance from the point $A(2,-1,1)$ is 6 units.

Given that $\boldsymbol{x}$-coordinate of $\mathbf{P}$ is greater than $\mathbf{2}$,

Find the coordinates of point Q on the line $L_2: x-1=\frac{y-2}{2}=\frac{z-2}{2}$ such that $\mathbf{Q}$ is the closest point to $\mathbf{P}$.

The measure of the angle between the lines $$x=k+1, \quad y=2 k-1, \quad z=2 k+3, \quad k \in R \quad$$ and $$\quad \frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-2}$$ is

The co-ordinate of the foot of the perpendicular from $$P(1,8,4)$$ on the line joining $$R(0,-1,3)$$ and $$Q(2,-3,-1)$$ is

If the straight lines $$\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-t}$$ and $$\frac{x-1}{t}=\frac{y-4}{2}=\frac{z-5}{1}$$ are intersecting then $$t$$ can have

If the line $$\frac{1-x}{-3}=y=\frac{z+2}{2}$$ is perpendicular to the line $$\frac{3 x-1}{2 b}=3-y=\frac{z-1}{a}$$, then find the value of $$3 a+3 b$$

The lines $$\vec{r}=(2 \hat{\jmath}-3 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})$$ and $$\vec{r}=(2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k})+\mu(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})$$ are

A line makes the same angle $$\theta$$ with each of the $$x$$ and $$z$$-axes. If the angle $$\beta$$, which it makes with the $$y$$-axis is such that $$\sin ^2 \beta=3 \sin ^2 \theta$$, then $$\cos ^2 \theta$$ equals

The foot of the perpendicular from $$(2,4,-1)$$ to the line $$x+5=\frac{1}{4}(y+3)=-\frac{1}{9}(z-6)$$ is

$$P$$ is a point on the line segment joining the points $$(3,2,-1)$$ and $$(6,2,-2)$$. If $$x$$ coordinate of $$\mathrm{P}$$ is 5, then its $$y$$ co-ordinate is

The vector equation of two lines are

$$\begin{aligned} & \vec{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k} \\ & \vec{r}=(s+1) \hat{\imath}+(2 s-1) \hat{\jmath}-(2 s+1) \hat{k} \end{aligned}$$

Then the shortest distance between them is

The place $$x-2 y+z=0$$ is parallel to the line

The lines $$\frac{x-1}{2}=\frac{y-4}{4}=\frac{z-2}{3}$$ and $$\frac{1-x}{1}=\frac{y-2}{5}=\frac{3-z}{a}$$ are perpendicular to each other, then $$a$$ equals to

If two lines $$L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$$ and $$L_2: \frac{x-3}{1}=\frac{y-k}{2}=z$$ intersect at a point, then $$2 k$$ is equal to

If the direction ratios of two lines are given by $$3 l m-4 l n+m n=0$$ and $$l+2 m+3 n=0$$, then the angle between the lines is

The coordinates of the vertices of the triangle are $$A(-2,3,6), B(-4,4,9)$$ and $$C(0,5,8)$$. The direction cosines of the median $$\mathrm{BE}$$ are

$$ \mathrm{P} \text { is a point on the line segment joining the points }(3,2,-1) \text { and }(6,2,-2) \text {. If the } x \text { co ordinate of } \mathrm{P} \text { is } 5 \text {, then its } \mathrm{y} \text { coordinate is } $$

The distance of the point $$(2,3,4)$$ from the line $$1-x=\frac{y}{2}=\frac{1}{3}(1+z)$$ is

If the position vector of a point $$A$$ is $$\vec{a}+2 \vec{b}$$ and $$\vec{a}$$ divides $$A B$$ in the ratio $$2: 3$$, then the position vector of $$B$$ is

The line $$\frac{x-3}{4}=\frac{y-4}{5}=\frac{z-5}{6}$$ is parallel to the plane

The angle between the lines $${{x + 4} \over 3} = {{y - 1} \over 5} = {{z + 3} \over 4}$$ and $${{x + 1} \over 1} = {{y - 4} \over 1} = {{z - 5} \over 2}$$ is

The point of intersection of the lines $${{x - 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$$ and $${{x -5} \over 2} = {{y - 2} \over 1} = z$$ is

The line $${{x - 2} \over 3} = {{y - 3} \over 4} = {{z - 4} \over 5}$$ is parallel to the plane

The equation of a plane passing through the line of intersection of the planes $$x+2y+3z=2,x-y+z=3$$ and at a distance $$\frac{2}{\sqrt3}$$ from the point $$(3,1,-1)$$ is

The angle between the lines $$2x=3y=-z$$ and $$6x=-y=-4z$$ is

The point of intersection of the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$ and $${{x - 4} \over 5} = {{y - 1} \over 2} = z$$ is

A vector perpendicular to the plane containing the points $$A(1, - 1,2),B(2,0, - 1),C(0,2,1)$$ is