Finding Plane Intercepts
To find the intercepts of the plane $3x - 4y - 2z = 6$ with the coordinate axes, we need to convert the equation into the intercept form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Here, $a$, $b$, and $c$ represent the x, y, and z intercepts, respectively.
Equation Conversion
First, divide the given equation $3x - 4y - 2z = 6$ by 6 to make the right-hand side equal to 1:
$ \frac{3x}{6} - \frac{4y}{6} - \frac{2z}{6} = \frac{6}{6} $
Simplify the fractions:
$ \frac{x}{2} - \frac{y}{\frac{6}{4}} - \frac{z}{\frac{6}{2}} = 1 $
$ \frac{x}{2} - \frac{y}{\frac{3}{2}} - \frac{z}{3} = 1 $
Identifying Intercepts
Rewrite the equation to match the standard intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. We adjust the signs in the denominators:
$ \frac{x}{2} + \frac{y}{-\frac{3}{2}} + \frac{z}{-3} = 1 $
By comparing this with the standard form, we can identify the intercepts:
- x-intercept ($a$): 2
- y-intercept ($b$): $-\frac{3}{2}$
- z-intercept ($c$): -3
Conclusion
The intercepts made by the plane $3x - 4y - 2z = 6$ with the coordinate axes are 2, $-\frac{3}{2}$, and -3.
Similar Questions
- Find the relation between x and y such that the point (x, y) is equidistant from (6, 2) and (4, 6).
- Find the area of a triangle formed by $(1, 0)$, $(-1, 0)$, $(0, 1)$.
- The points A (1, 2), B (3, 4) and C (4, 1) are the vertices of a triangle which is:
- Three straight lines $x + y - 3 = 0$, $x + y + 2 = 0$ and $3x + 3y - 7 = 0$ are:
- The image of the point $(7, 8)$ when reflected along the x-axis is:
- The equation of a straight line passing through (-2,5) and (1,3) is:
- Find the relation between x and y such that the point (x, y) is equidistant from (5, 3) and (4, 6).
- The distance between two points $(a \cos \alpha, 0)$ and $(0, a \sin \alpha)$ is_____.
- The area of the triangle formed by the line $2x - 4y - 7 = 0$ with the coordinate axis is:
- The distance from the origin to the line $4x + 3y + 6 = 0$ is:
Important Questions from Coordinate Geometry
What is the reflection of the point (-1, 5) in the line x = 1?
What are the co-ordinates of the centroid of a triangle, whose vertices are A(1, -5), B(-4, 0) and C(3, -4)?
Slope of the line AB is 4/3. Co-ordinates of points A and B are (x, -5) and (2, -3) respectively. What is the value of x?
Find the co-ordinates of the centroid of a triangle whose vertices are A(1, 4), B(7, 8) and C(10, 12).
- If x² + y² - 12x + 18y + 117 = 0, then the value of x² + y² is:
