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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

For \(\mathrm{A}(7,-3,1)\) and \(\mathrm{B}(4,9,8)\), the point that divides \(\underline{\mathrm{AB}}\) from \(\mathrm{B}\) in the ratio \(2: 5\) is (A) \([(34 / 7),(39 / 7 \overline{),(42} / 7)]\) (B) \([(34 / 7),(39 / 7),\\{(-42) / 7\\}]\) (C) \([\\{(-34) / 7\\},(39 / 7),\\{(-42) / 7\\}]\) (D) \([\\{(-34) / 7\\},\\{(-39) / 7\\},\\{(-42) / 7\\}]\)

\([(2-3 \mathrm{x}) / 6]=[(\mathrm{y}+1) / 2]=[(1-z) /(-2)]\) direction of line \(\overline{(A)}+2,2,2\) (B) \(-1,1,1\) (C) \(-3,2,2\) (D) \(6,2,-2\)

Lines \(\underline{r}=(1,3,5)+k(-1,2,3), k \in R\) and \(\underline{r}=(1,3,1)+\mathrm{k}(1,-2,3), \mathrm{k} \in \mathrm{R}\) are (A) coincident (B) parallel (C) skew (D) perpendicular

If \((\underline{a}+\underline{b}) \cdot(\underline{a}-\underline{b})=63\) and \(|\underline{a}|=8|\underline{b}|\), then \(|\underline{a}|=\) (A) 8 (B) 64 (C) 16 4

If vector \(\underline{x}\) forms an equal angle \(\alpha\) with three axis and \(|\underline{x}|=9\) then \(\alpha=\quad\) where \(0<\alpha<(\pi / 2)\) (A) \(\cos ^{-1}(1 / \sqrt{2})\) (B) \(\cos ^{-1}(1 / 9)\) (C) \(\cos ^{-1}(1 / \sqrt{3})\) (D) \(\cos ^{-1}(1 / 3)\)
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