In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z=2 (b) x+y+z=1
(c)2x+3y-z=5 (d) 5y+8=0
(a) The equation of the plane is z=2 or 0x+0y+z=2...(1)
The direction ratios of the normal are 0, 0, and 1.
∴\(\sqrt {0^2+0^2+1^2}=1\)
Dividing both sides of equation(1) by 1, we obtain
0.x+0.y+1.z=2
This is of the form lx+my+nz=d, where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.
Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.
(b) x+y+z=1...(1)
The direction ratios of normal are 1, 1, and 1.
∴\(\sqrt {(1)^2+(1)^2+(1)^2}=\sqrt 3\)
Dividing both sides of equation(1) by \(\sqrt 3\), we obtain
\(\frac{1}{\sqrt 3}x+\frac{1}{\sqrt 3}y+\frac{1}{\sqrt 3}z=\frac{1}{\sqrt 3}\) ...(2)
This equation is the form lx+my+nz=d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal are \(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\), and \(\frac{1}{\sqrt 3}\) and the distance of normal from the origin is \(\frac{1}{\sqrt 3}\) units.
(c) 2x+3y-z=5...(1)
The direction ratios of normal are 2, 3, and -1.
∴\(\sqrt{(2)^2+(3)^2+(-1)^2}=\sqrt 14\)
Dividing both sides of equation(1) by 14, we obtain
\(\frac{2}{\sqrt {14}}x+\frac{3}{\sqrt{14}}y-\frac{1}{\sqrt {14}}z=\frac{5}{\sqrt {14}}\)
This equation is of the form lx+my+nz=d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are \(\frac{2}{\sqrt {14}},\frac{3}{\sqrt {14}}\), and \(\frac{-1}{\sqrt{14}}\) and the distance of normal from the origin is \(\frac{5}{\sqrt{14}}\) units.
(d) 5y+8=0
\(\Rightarrow \) 0x-5y+0z=8...(1)
The direction ratios of normal are 0, -5, and 0.
∴\(\sqrt{0+(-5)^2+0}\) =5
Dividing both sides of equation(1) by 5, we obtain
-y=\(\frac{8}{5}\)
This equation is of the form lx+my+nz=d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are 0, -1, and 0 and the distance of normal from the origin is \(\frac{8}{5}\) units.
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
The vector equations of two lines are given as:
Line 1: \[ \vec{r}_1 = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(4\hat{i} + 6\hat{j} + 12\hat{k}) \]
Line 2: \[ \vec{r}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(6\hat{i} + 9\hat{j} + 18\hat{k}) \]
Determine whether the lines are parallel, intersecting, skew, or coincident. If they are not coincident, find the shortest distance between them.
Determine the vector equation of the line that passes through the point \( (1, 2, -3) \) and is perpendicular to both of the following lines:
\[ \frac{x - 8}{3} = \frac{y + 16}{7} = \frac{z - 10}{-16} \quad \text{and} \quad \frac{x - 15}{3} = \frac{y - 29}{-8} = \frac{z - 5}{-5} \]
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely: