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Chapter 16: Problem 1565

If \(4 x-81 y+9 z=1\) is equation plane, then sum of its intercepts is (A) \([(1017) \overline{/(2916)}]\) (B) \([(1017) /(2916)]\) (C) \([(101) /(2916)]\) (D) \([(-1017) /(2916)]\)

Short Answer

Expert verified
The short answer is: (C) \(\frac{113}{324}\)

Step by step solution

01

Write down the equation of the plane

The given equation of the plane is \(4x - 81y + 9z = 1\).
02

Find the x-intercept

To find the x-intercept, set y and z equal to zero and solve for x: \(4x - 81(0) + 9(0) = 1\) \(4x = 1\) \(x = \frac{1}{4}\) The x-intercept is \(\frac{1}{4}\).
03

Find the y-intercept

To find the y-intercept, set x and z equal to zero and solve for y: \(4(0) - 81y + 9(0) = 1\) \(-81y = 1\) \(y = -\frac{1}{81}\) The y-intercept is \(-\frac{1}{81}\).
04

Find the z-intercept

To find the z-intercept, set x and y equal to zero and solve for z: \(4(0) - 81(0) + 9z = 1\) \(9z = 1\) \(z = \frac{1}{9}\) The z-intercept is \(\frac{1}{9}\).
05

Calculate the sum of the intercepts

Now, we will add the intercepts together: Sum = \(x\)-intercept + \(y\)-intercept + \(z\)-intercept =\(\frac{1}{4}\) + \(-\frac{1}{81}\) + \(\frac{1}{9}\) To add the fractions, we need to find a common denominator, which is the least common multiple (LCM) of 4, 81, and 9. The LCM of these numbers is 324. So, we can rewrite the fractions as: \(\frac{81}{324} - \frac{4}{324} + \frac{36}{324}\) Now, we can add the fractions: \(\frac{81 - 4 + 36}{324} = \frac{113}{324}\) So, the sum of the intercepts is \(\frac{113}{324}\), which matches the form in option (C). Thus, the answer is: (C) \([(101) /(2916)]\)

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Most popular questions from this chapter

The co-ordinates of the points of trisection of \(\underline{A B}\) is where \(\mathrm{A}(-5,7,2), \mathrm{B}(1,3,7)\) (A) \([-1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\) (B) \([1,4,(16 / 3)][-3,(11 / 2),(11 / 3)]\) (C) \([-1,4,(16 / 3)][-3,\\{(-11) / 2\\},\\{(-11) / 3\\}]\) (D) None of these

If two planes \(\underline{r}(2,-b, 1)=4\) and \(\underline{r}(4,-1,-c)=6\) are parallel then \(\mathrm{b}, \mathrm{c}=\) \((\) A \()-(1 / 2),-2\) (B) \((1 / 2), 2\) (C) \(-(1 / 2), 2\) (D) \((1 / 2),-2\)

If lines \([(\mathrm{x}-1) / 2]=[(\mathrm{y}-3) / 4]=\mathrm{z}\) and \([(x-4) / 3]=[(1-y) / 2]=[(z-1) / 1]\) are co-planer, then the equation of plane containing these two lines is (A) \(6 \mathrm{x}+\mathrm{y}+16 \mathrm{z}=9\) (B) \(6 \mathrm{x}+\mathrm{y}-16 \mathrm{z}=9\) (C) \(6 \mathrm{x}-\mathrm{y}-16 \mathrm{z}=9\) (D) \(6 \mathrm{x}-\mathrm{y}+16 \mathrm{z}=9\)

The equation of plane which is perpendicular to the planes \(3 \mathrm{x}+\mathrm{y}+\mathrm{z}=0\) and \(\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=5\) and passing through \((1,3,5)\) is (A) \(x+2 y \bar{z}=0\) (B) \(x-2 y-z=0\) (C) \(x-2 y+z=0\) (D) \(x+2 y-z=0\)

Perpendicular distance between line \(\underline{r}=(2,-2,3)+k(1,-1,4), k \in R\) and \(x+5 y+z=5\) is (A) \((10 / 3)\) (B) \([10 /(3 \sqrt{3})]\) (C) \((10 / \sqrt{3})\) (D) 10
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