Equation of a Plane : Point Normal Form

Q.

Find the direction ratios of the plane $\frac{(3-x)}{1}=\frac{(y-2)}{5}=\frac{(2z-3)}{1}$.

1. $1,5,\frac{1}{2}$

2. $-1,5,1$

3. $-1,5,\frac{1}{2}$

4. $1,5,1$

Q. The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :

1. 125x – 90y – 79z = 340
2. 32x – 21y – 36z = 85
3. 73x + 61y – 22z = 85
4. 29x – 27y – 22z = 85

Q. The projection of the line $\frac{x-2}{4}=\frac{y+2}{8}=\frac{z-3}{6}$ in the plane $P_1:2x+3y-4z+6=0$ is the line on the intersection of planes $P_1$ and $P_2$. If $P_1$ is perpendicular to $P_2$, then equation of the plane $P_2$ is

1. $25x-14y+2z=84$
2. $25x-14y+2z=-34$
3. $-25x-14y+2z=84$
4. $25x+14y+2z=34$

Q. Let the projection of the line $L:\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z-2}{3}$ in the plane $3x+2y+z=5$ is $L^{\prime }$. If the distance of a plane from the origin which contains the lines $L$ and $L^{\prime }$ is $p$, then the value of $6p^2$ is

Q. Let the projection of the line $L:\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z-2}{3}$ in the plane $3x+2y+z=5$ is $L^{\prime }$. If the distance of a plane from the origin which contains the lines $L$ and $L^{\prime }$ is $p$, then the value of $6p^2$ is

Q.

Consider the lines
$L_1:\frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1},L_2:\frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2},$ and the planes $P_1:7x+y+2z=3,P_2:3x+5y-6z=4.$ Let $ax+by+cz=d$ the equation of the plane passing through the point of intersection of lines $L_1$ and $L_2$ and perpendicular to planes $P_1$ and $P_2$.
Match List I with List II and select the correct answer using the code given below the lists.

$$\begin{array}{cccc}ListI& & ListII& \\ P& a=& 1& 13\\ Q& b=& 2& -3\\ R& c=& 3& 1\\ S& d=& 4& -2\end{array}$$

1. P-3, Q-2, R-4, S-1

2. P-1, Q-3, R-4, S-2

3. P-3, Q-2, R-1, S-4

4. P-2, Q-4, R-1, S-3

Q.

Consider the lines
$L_1:\frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1},L_2:\frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2},$ and the planes $P_1:7x+y+2z=3,P_2:3x+5y-6z=4.$ Let $ax+by+cz=d$ the equation of the plane passing through the point of intersection of lines $L_1$ and $L_2$ and perpendicular to planes $P_1$ and $P_2$.
Match List I with List II and select the correct answer using the code given below the lists.

$$\begin{array}{cccc}ListI& & ListII& \\ P& a=& 1& 13\\ Q& b=& 2& -3\\ R& c=& 3& 1\\ S& d=& 4& -2\end{array}$$

1. P-3, Q-2, R-4, S-1

2. P-1, Q-3, R-4, S-2

3. P-3, Q-2, R-1, S-4

4. P-2, Q-4, R-1, S-3

Q. The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :

1. 125x – 90y – 79z = 340
2. 32x – 21y – 36z = 85
3. 73x + 61y – 22z = 85
4. 29x – 27y – 22z = 85

Q. Equation of the plane perpendicular to the plane x – 2y + 5z + 1 = 0 which passes through the points (2, –3, 1) and (–1, 1, –7) is given by :

1. 4x – 4y + z + 7 = 0
2. 4x + 7y + 2z + 11 = 0
3. 2x + y – z = 0
4. 2x + y – 3z = 0

Q.

The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1) having normal perpendicular to both the lines
$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$, is

1. $\frac{20}{\sqrt{74}}$ units

2. $\frac{10}{\sqrt{83}}$ units

3. $\frac{5}{\sqrt{83}}$ units

4. $\frac{10}{\sqrt{74}}$ units