Given: Equation of plane is \(\bar{r}\)(2\(\hat{i}\) - 3\(\hat{j}\) + 4\(\hat{k}\)) = 12
To find: Intercepts made by the plane.
Formula Used: Equation of plane is \(\frac{x}a\) + \(\frac{y}b\) + \(\frac{z}c\) = 1 where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively.
Explanation:
The equation of the plane can be written as
2x – 3y + 4z = 12
Dividing by 12,
\(\frac{x}6\) + \(\frac{y}{-4}\) + \(\frac{z}3\) = 1 which is of the form \(\frac{x}a\) + \(\frac{y}b\) + \(\frac{z}c\) = 1
Therefore the intercepts made by the plane are 6, -4, 3
