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Chapter 16: Problem 1577

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) \((1 / 2),-2\)

Short Answer

Expert verified
(D) (1/2, -1)

Step by step solution

01

Write the planes in general form

We start by writing the given planes in their general form which is \(Ax + By + Cz = D\): Plane 1: \(2x - by + z = 4\) Plane 2: \(4x - y - cz = 6\)
02

Find the normal vectors for each plane

The normal vector of a plane can be found using the coefficients of x, y, and z in the general form of the equation. For the given planes, the normal vectors are: Normal vector of Plane 1: \(\underline{n_1} = \langle 2, -b, 1 \rangle\) Normal vector of Plane 2: \(\underline{n_2} = \langle 4, -1, -c \rangle\)
03

Check if the normal vectors are proportional

Two normal vectors are proportional (parallel) if there is a constant scalar k such that \(\underline{n_1} = k \cdot \underline{n_2}\). That is, for \(\underline{n_1}\) and \(\underline{n_2}\) to be proportional: \(\frac{2}{4}=\frac{-b}{-1}=\frac{1}{-c}\) Now, we will solve these equations to find values of b and c.
04

Determine the values of b and c

From the proportionality between \(\underline{n_1}\) and \(\underline{n_2}\), we can write the following equations: \(k=\frac{2}{4} \Rightarrow k=\frac{1}{2}\) \(k=\frac{-b}{-1} \Rightarrow -b = -\frac{1}{2} \) \(k=\frac{1}{-c} \Rightarrow -c=1\) Now, solve for b using the second equation: \(-b = -\frac{1}{2} \Rightarrow b = \frac{1}{2}\) And for c using the third equation: \(-c = 1 \Rightarrow c =-1\) Therefore, we have: b = 1/2, c = -1 Matching this with the given options, we get the correct answer is: (D) (1/2, -1)

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Most popular questions from this chapter

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

shortest distance between two lines \(\underline{\mathrm{r}}=(4,-1,0)+\mathrm{k}(1,2,-3), \mathrm{k} \in \mathrm{R}\) and \(\underline{\mathrm{r}}=(1,-1,2)+\mathrm{k}(2,4,-5), \mathrm{k} \in \mathrm{R}\) is (A) \((6 / \sqrt{5})\) (B) \((6 / 5)\) (C) \((\sqrt{6} / 5)\) (D) \(\sqrt{(6 / 5)}\)

Direction cosine of line \(\mathrm{x}=3-2 \mathrm{y}, \mathrm{z}=2 \mathrm{y}-1\) is (A) \([\\{(-2) / 3\\},(1 / 3),(2 / 3)]\) (B) \([\\{(-2) / 3\\},\\{(-1) / 3\\},(2 / 3)]\) (C) \([(2 / 3),(1 / 3),(2 / 3)]\) (D) None of these

\(\mathrm{P}(1,6,3)\) to \((\mathrm{x} / 1)=[(\mathrm{y}-1) / 2]=[(\mathrm{z}-2) / 3]\) on then image of \(\mathrm{P}\) is (A) \(\overline{(-1,0,-7)}\) (B) \((-1,0,7)\) (C) \((1,0,7)\) (D) \((1,0,-7)\)

The equation of plane passing through the lines \((\mathrm{x} / 2)=[(\mathrm{y}-1) / 1]=[(\mathrm{z}+2) / 2]\) and \([(2 \mathrm{x}+3) / 4]=[(3-\mathrm{y}) /(-1)]=(\mathrm{z} / 2)\) is (A) \(4 \mathrm{x}+11 \mathrm{y}+14 z=36\) (B) \(\overline{4 x+1} 4 y-11 z=36\) (C) \(4 \mathrm{x}-14 \mathrm{y}-11 \mathrm{z}=36\) (D) \(4 \mathrm{x}-14 \mathrm{y}+11 \mathrm{z}=36\)
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