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Chapter 16: Problem 1511

For vector a and \(\underline{b},|\underline{a}+\underline{b}|<|\underline{a}-\underline{b}|\), then the angle between \(\underline{a}\) and \(\underline{b}\) is (A) obtuse (B) Acute (C) Right (D) supplementary

Short Answer

Expert verified
The angle between the vectors \(\underline{a}\) and \(\underline{b}\) is obtuse.

Step by step solution

01

Write down given inequality

We are given the inequality: \[|\underline{a} + \underline{b}| < |\underline{a} - \underline{b}|\]
02

Use dot product properties

Recall that for any two vectors \(\underline{v}\) and \(\underline{w}\), \(|\underline{v}|^2 = \underline{v} \cdot \underline{v}\). Now let's square both sides of the inequality: \[|(\underline{a}+\underline{b})|^2<|(\underline{a}-\underline{b})|^2\] Using the properties of dot products and expanding on both sides of the inequality, we have: \[(\underline{a}+\underline{b}) \cdot (\underline{a}+\underline{b})<(\underline{a}-\underline{b})\cdot(\underline{a}-\underline{b})\]
03

Apply the distributive property of dot products

Applying the distributive property of dot products on both sides, we get: \[\underline{a}\cdot\underline{a}+2(\underline{a}\cdot\underline{b})+\underline{b}\cdot\underline{b}<\underline{a}\cdot\underline{a}-2(\underline{a}\cdot\underline{b})+\underline{b}\cdot\underline{b}\]
04

Cancelling and simplifying terms

Now let's cancel and simplify the terms in the expression: \[4(\underline{a}\cdot\underline{b})<0\] Divide both sides by 4: \[\underline{a} \cdot \underline{b} < 0\]
05

Using the relationship between dot products and angles

The dot product of two vectors \(\underline{a}\) and \(\underline{b}\) can be expressed in terms of their magnitudes and the angle between them: \[\underline{a}\cdot\underline{b}=|\underline{a}||\underline{b}|\cos\theta\] Since \(\underline{a}\cdot\underline{b}<0\), this implies that \(\cos\theta<0\). We know that cosine is negative in the second and third quadrants (from \(90^\circ\) to \(270^\circ\)). This means that the angle between these two vectors is an obtuse angle. The correct choice is: (A) obtuse

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Most popular questions from this chapter

If \(\mathrm{A}(\mathrm{a}, 1,3), \mathrm{B}(-1, \mathrm{~b}, 2), \mathrm{C}(1,0, \mathrm{c})\) are the vertices of \(\Delta \mathrm{ABC}\) whose centroid is \((2,3,5)\), then values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are respectively (A) \(10,8,6\) (B) \(6,10,8\) (C) \(8,6,10\) (D) \(6,8,10\)

For vectors \(\underline{a}, \underline{b}, c \underline{c}\) if each vector is perpendicular to the sum of remaining two vectors and \(|\underline{\mathrm{a}}|=3,|\underline{\mathrm{b}}|=4,|\underline{\mathrm{c}}|=5\), then \(|\underline{a}+\underline{b}+\underline{c}|=\) (A) \(2 \sqrt{2}\) (B) \(3 \sqrt{2}\) (C) \(4 \sqrt{2}\) (D) \(5 \sqrt{2}\)

If \(\mathrm{m} \angle \mathrm{B}=(\pi / 2)\) in \(\Delta \mathrm{ABC}\) and \(\mathrm{P}, \mathrm{Q}\) are points of trisection of hypotenuse \(\underline{\mathrm{A}} \mathrm{C}\), then \(\mathrm{BP}^{2}+\mathrm{BQ}^{2}=\) (A) \((5 / 9) \mathrm{AC}^{2}\) (B) \((5 / 9) \mathrm{AC}\) (C) \((25 / 81) \mathrm{AC}^{2}\) (D) \((25 / 81) \mathrm{AC}\)

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) \(\cos ^{-1}(1 / 3)\)

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) \([(-1017) /(2916)]\)
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