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Chapter 17: Problem 1664

if the angle between \(\underline{\mathrm{a}}=(2,-\mathrm{m}, 3 \mathrm{~m})\) and \(\underline{\mathrm{b}}=(1+\mathrm{m},-2 \mathrm{~m}, 1)\) is acute then \(\mathrm{m}=\) (a) \(\mathrm{m} \in \mathrm{R}\) (b) \(\mathrm{m} \in \overline{(-\infty,}-2) \cup[-(1 / 2), \infty]\) (c) \(\mathrm{m}=-(1 / 2)\) (d) \(m \in[-2,-(1 / 2)]\)

Short Answer

Expert verified
\(m \in \left(-\infty, -\frac{1}{2}\right) \cup (2, \infty)\)

Step by step solution

01

Define the vectors and their dot product

The problem gives us the vectors \(\vec{a}=(2, -m, 3m)\) and \(\vec{b}=(1+m, -2m, 1)\). We begin by computing their dot product. The dot product of two vectors is defined as: $$ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3, $$ where \(a_1, a_2,\) and \(a_3\) are the components of \(\vec{a}\) and \(b_1, b_2,\) and \(b_3\) are the components of \(\vec{b}\).
02

Compute the dot product

Using the components of the given vectors, we find their dot product: $$ \vec{a} \cdot \vec{b} = (2)(1+m) + (-m)(-2m) + (3m)(1) = 2 + 2m - 2m^2 + 3m. $$
03

Apply the condition for an acute angle

The angle between two vectors is acute if their dot product is positive. In other words: $$ \vec{a} \cdot \vec{b} > 0. $$ Now, let's substitute the computed dot product for \(\vec{a} \cdot \vec{b}\): $$ 2 + 2m - 2m^2 + 3m > 0. $$
04

Simplify and solve the inequality

Next, we simplify the inequality: $$ 2 - 2m^2 + 5m > 0. $$ Now, we need to find the roots of the quadratic and check the interval in between: $$ m_1 = -(\frac{1}{2}) \quad \text{and} \quad m_2 = 2. $$ Hence the solution set for \(m\) is: $$ m \in \left(-\infty, -\frac{1}{2}\right) \cup (2, \infty). $$ So the correct answer is (b) \(m \in \left(-\infty, -\frac{1}{2}\right) \cup [-(1 / 2), \infty]\).

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Most popular questions from this chapter

\(\underline{\mathrm{a}}=(1,-1,0), \underline{\mathrm{b}}=(0,1,-1)\) and \(\underline{\mathrm{c}}=(-1,0,1)\) find the unit vector \(\underline{d}\) Such that \(\underline{\mathrm{a}} \cdot \underline{\mathrm{d}}=0=[\underline{\mathrm{b}} \underline{\mathrm{c}} \underline{\mathrm{d}}]\). (a) \(\pm(1 / \sqrt{6})(1,1,-2)\) (b) \(\pm(1 / \sqrt{3})(1,1,-1)\) (c) \(\pm(1 / \sqrt{3})(1,1,1)\) (d) \(\pm(0,0,1)\)

if \(\underline{u}\) and \(\underline{\mathrm{v}}\) are unit vectors and \(\theta\) is acute angle between them find \(\theta\) such that \(2 \underline{\mathrm{u}} \times 3 \underline{\mathrm{v}}\) becomes unit vectors. (a) \(\sin ^{-1}(1 / 3)\) (b) \(\sin ^{-1}(1 / 6)\) (c) \(\cos ^{-1}(1 / 6)\) (d) none of these

\(\underline{\mathrm{a}}=(1,0,-1), \underline{\mathrm{b}}=(\mathrm{x}, 1,1-\mathrm{x})\) and \(\underline{\mathrm{c}}=(\mathrm{y}, \mathrm{x}, 1+\mathrm{x}-\mathrm{y})\), \([\underline{\mathrm{a}} \underline{\mathrm{b}} \underline{\mathrm{c}}]\) is depend on which. (a) \(x\) (b) \(\mathrm{y}\) (c) \(x\) and \(y\) (d) none of these

\((1 / \mathrm{c}),(1 / \mathrm{c}),(1 / \mathrm{c})\) is direction cosine of the line then the value of \(\mathrm{c}=\) (a) \(\pm(\overline{1 / 3)}\) (b) \(\pm 3\) (c) \(\pm(1 / \sqrt{3})\) (d) \(\pm \sqrt{3}\)

find the vector in \(\mathrm{R}^{2}\) such as perpendicular with \(\underline{\mathrm{x}}=(3,4)\) as well as acute angle with Y-axes. (a) \([(4 / 5),(3 / 5)]\) (b) \([-(4 / 5),(3 / 5)]\) (c) \([-(4 / 5),-(3 / 5)]\) (d) \([(3 / 5),-(4 / 5)]\)
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