Equation of Line: Symmetrical Form

Q. The equation of the line passing through the point (1, 1, –1) and perpendicular to the plane x – 2y – 3z = 7 is

1. $\frac{x-1}{-1}=\frac{y-1}{2}=\frac{z+1}{3}$
2. $\frac{x-1}{-1}=\frac{y-1}{-2}=\frac{z+1}{3}$
3. $\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z+1}{-3}$
4. $\frac{x+1}{1}=\frac{y+1}{-2}=\frac{z+1}{-3}$

Q. The equation of line $AB$ is $\frac{x}{2}=\frac{y}{-3}=\frac{z}{6}$. Through a point $P(1,2,5)$, line $PN$ is drawn perpendicular to $AB$ and line $PQ$ is drawn parallel to the plane $3x+4x+5z=0$ to meet $AB$ at $Q$. Then

1. Coordinates of $N$ are $(\frac{52}{49},-\frac{78}{49},\frac{156}{49})$
2. Equation of line $NQ$ is $3(x-3)=-(2y+9)=z-9$
3. Equation of line $NQ$ is $3(x-3)=(2y+9)=z-9$
4. Coordinates of $Q$ are $(3,\frac{9}{2},9)$

Q. If the image of the point $P(1,-2,3)$ in the plane, $2x+3y-4z+22=0$ measured parallel to the line, $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $Q,$ then $PQ$ is equal to

1. $3\sqrt{5}$
2. $2\sqrt{42}$
3. $\sqrt{42}$
4. $6\sqrt{5}$

Q. Two lines $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$ and $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$ intersect at the point $R$. The reflection of $R$ in the $xy$- plane has coordinates :

1. $(2,-4,-7)$
2. $(2,4,7)$
3. $(2,-4,-7)$
4. $(-2,4,7)$

Q. 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 =

1. $\frac{2}{9}$
2. $\frac{9}{2}$
3. \N
4. 3

Q. Equation of a line passing through (2, -1, 1) and parallel to the line whose equation is $\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3}$is

1. $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-1}{2}$
2. $\frac{x-2}{2}=\frac{y+1}{7}=\frac{z-1}{-3}$
3. $\frac{x-2}{2}=\frac{y-1}{-1}=\frac{z+3}{1}$
4. $\frac{x-3}{2}=\frac{y+1}{-1}=\frac{z-2}{1}$

Q. Equation of the line drawn through the point $(1,0,2)$ and perpendicular to the line $\frac{x+1}{3}=\frac{y-2}{-2}=\frac{z+1}{-1},$ is

1. $\frac{x-1}{2}=\frac{y}{-1}=\frac{z-2}{7}$
2. $\frac{x-1}{1}=\frac{y}{2}=\frac{z-2}{7}$
3. $\frac{x-1}{1}=\frac{y}{-2}=\frac{z-2}{7}$
4. $\frac{x-1}{2}=\frac{y}{1}=\frac{z-2}{-7}$

Q.

The equation of the line which passes through the point (1, 1, 1) and intersecting the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ is

1. $\frac{x-1}{4}=\frac{y-1}{11}=\frac{z-1}{13}$

2. $\frac{x-1}{17}=\frac{y-1}{-3}=\frac{z-1}{17}$

3. $\frac{x-1}{13}=\frac{y-1}{5}=\frac{z-1}{-2}$

4. $\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}$

Q.

What is an equation of the line that passes through the points in the table?

1. $\mathrm{y}=\frac{1}{3}\mathrm{x}+1$

2. $\mathrm{y}=4\mathrm{x}$

3. $\mathrm{y}=3\mathrm{x}+1$

4. $\mathrm{y}=\frac{1}{4}\mathrm{x}$

Q. Two lines $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$ and $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$ intersect at the point $R$. The reflection of $R$ in the $xy$- plane has coordinates :

1. $(2,-4,-7)$
2. $(2,4,7)$
3. $(2,-4,-7)$
4. $(-2,4,7)$

Q.

The distance of the point (1, 0, 2) from the point of intersection of the line
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane x - y + z = 16 is

1. $2\sqrt{14}$

2. 8

3. $3\sqrt{21}$

4. 13

Q. The point on the line $\frac{x-2}{1}=\frac{y+3}{-2}=\frac{z-5}{-2}=r\left(say\right)$ at a distance of 6 units from the point (2, –3, 5) is

1. (3, –5, –3)
2. (4, –7, –9)
3. (0, 2, –1)
4. (3, 2, 1)