Find the intercepts cut off by the plane 2x+y-z = 5
2x+y-z=5...(1)
Dividing both sides of equation(1) by 5, we obtain
\(\frac{2}{5}x+\frac{y}{5}-\frac{z}{5}=1\)
\(\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1\)...(2)
It is known that the equation of a plane in intercept form is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), where a,b,c are the intercepts cut off by the planet at x, y, and z axes respectively.
Therefore, for the given equation, a=\(\frac{5}{2}\), b=5, and c=-5
Thus, the intercepts cut off by the plane are \(\frac{5}{2}\), 5, and -5.
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: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
Information Table
| Information | Amount (₹) |
|---|---|
| Preference Share Capital | 8,00,000 |
| Equity Share Capital | 12,00,000 |
| General Reserve | 2,00,000 |
| Balance in Statement of Profit and Loss | 6,00,000 |
| 15% Debentures | 4,00,000 |
| 12% Loan | 4,00,000 |
| Revenue from Operations | 72,00,000 |
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: