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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

If perpendicular distance from \((0,0,0)\) to the variable plane is \(\mathrm{p}\) and variable plane intersects the axis in \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), the centroid of \(\Delta \mathrm{ABC}\) is on \(\left(1 / \mathrm{x}^{2}\right)+\left(1 / \mathrm{y}^{2}\right)+\left(1 / \mathrm{z}^{2}\right)=\) (A) \(\left(9 / \mathrm{p}^{2}\right)\) (B) \(\left(\mathrm{p}^{2} / 9\right)\) (C) \((\mathrm{p} / 9)\) (D) \((9 / \mathrm{p})\)

The foot of the perpendicular and perpendicular distance from point \((1,2,3)\) to plane \(x-2 y+2 z=5\) is and respectively (A) \([(11 / 9),(14 / 9),(31 / 9)],(2 / 3)\) (B) \([\\{(-11) / 9\\},\\{(-14) / 9\\},\\{(-31) / 9\\}],(2 / 3)\) (C) \([(11 / 9),(14 / 9),(31 / 9)],(2 / 3)\) (D) \([(11 / 9),(14 / 9),\\{(-31) / 9\\}],(2 / 3)\)

Line \([(x-1) / c]=[(y+3) /(-1)]=[(z-3) / 2]\) and \([(x-3) / 6]=[(y-1) / 3]=[(4-z) / 6]\) if direction are same then \(\mathrm{c}=\) (A) \(-2\) (B) 2 (C) \((1 / 3)\) (D) \(-(1 / 3)\)

If the vertices of quadrilateral are \((1,1,1),(-2,4,1)\) \((-1,5,5),(2,2,5)\) then it is (A) rectangle (B) square (C) parallelogram (D) rhombus

If angle between \(\underline{a}\) and \(\underline{b}\) is \(\theta\), then \([(|\underline{\mathbf{a}} \times \underline{\mathrm{b}}|) /(\underline{\mathrm{a}} \cdot \underline{\mathrm{b}})]=\) \(\begin{array}{llll}(\text { A })-\cot \theta & \text { (B) }-\tan \theta & \text { (C) } \tan \theta & \text { (D) } \cot \theta\end{array}\)
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