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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

Line of intersection of the planes \(2 \mathrm{x}+\mathrm{y}+2 \mathrm{z}=1\), \(x+2 y-2 z=1\) and \(6 x+2 y+3 z=1,6 x+2 y-3 z=1\) is and point of intersection is (A) intersecting \((1,1,1)\) (B) Perpendicular \((-1,1,1)\) (C) non-co-planer lines, does not exist (D) Parallel, does not exist

The equation of plane which is perpendicular to the planes \(3 \mathrm{x}+\mathrm{y}+\mathrm{z}=0\) and \(\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=5\) and passing through \((1,3,5)\) is (A) \(x+2 y \bar{z}=0\) (B) \(x-2 y-z=0\) (C) \(x-2 y+z=0\) (D) \(x+2 y-z=0\)

The angle between two unit vectors a and \(\underline{b}\) is \(\theta,|\underline{a}+\underline{b}|<1\) if (A) \(\theta=(\pi / 2)\) (B) \(\theta>(\pi / 3)\) (C) \((2 \pi / 3)<\theta<\pi\) (D) \(\theta=(\pi / 6)\)

The equation of the plane which intersects the axis at \(\mathrm{A}, \mathrm{B}\), \(\mathrm{C}\) and the centroid of \(\Delta \mathrm{ABC}\) is \((\alpha, \beta, \gamma)\) is (A) \(x+y+z=3 \alpha \beta \gamma\) (B) \(x+y+z=3\) (C) \((\mathrm{x} / \alpha)+(\mathrm{y} / \beta)+(\mathrm{z} / \gamma)=3\) (D) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=\alpha \beta \gamma\)

The direction cosine of \(\mathrm{x}=\mathrm{ay}+\mathrm{b}, \mathrm{z}=\mathrm{cy}+\mathrm{d}\) (A) \(\pm\left[a / \sqrt{\left(a^{2}+c^{2}+1\right)}\right], \pm\left[1 / \sqrt{ \left.\left(a^{2}+c^{2}+1\right)\right]}\right.\) \(\left.\pm\left[\mathrm{c} / \sqrt{(} \mathrm{a}^{2}+\mathrm{c}^{2}+1\right)\right]\) (C) \(\left[(-a) / \sqrt{\left(a^{2}+c^{2}+1\right)}\right],\left[(-1) / \sqrt{ \left.\left(a^{2}+c^{2}+1\right)\right]}\right.\) \(\left[(-c) / \sqrt{\left(a^{2}+c^{2}+1\right)}\right]\) (D) None of these
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