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The intercepts made by the plane \(\bar{r}\)(2\(\hat{i}\) - 3\(\hat{j}\) + 4\(\hat{k}\)) = 12 areA. 2, -3, 4 B. 2, -3, -6 C. 6, -4, 3 D. -6, 4, 3 |
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Answer» 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 |
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