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Chapter 17: Problem 1604

$$ \begin{aligned} &\underline{x} \text { and } y \text { are nonzero vector. If } \underline{x}=k y, k<0 \text { then } \underline{x} \cdot y=\\\ &\text { (a) }=|\underline{x}+y| \quad(b)=|\underline{x}||y| \quad \text { (c) }>|\underline{x}||y|\\\ &\text { (d) }<|\underline{x}| \mid y \end{aligned} $$

Short Answer

Expert verified
The correct option is (d), \(\underline{x} \cdot y < |\underline{x}||y|\), since \(k < 0\) and the dot product of \(\underline{x}\) and \(y\) will be negative, while the product of the magnitudes will be positive.

Step by step solution

01

Rewrite the expression for \(\underline{x}\)

Given that \(\underline{x} = ky\), where \(k < 0\), we can substitute this expression into the dot product equation.
02

Compute the dot product of \(\underline{x}\) and \(y\)

We know that the dot product of two vectors \(\underline{a}\) and \(\underline{b}\) is given by \(|\underline{a}||\underline{b}|\cos(\theta)\), where \(\theta\) is the angle between the two vectors. We can rewrite the dot product of \(\underline{x}\) and \(y\) using the expression for \(\underline{x}\) from Step 1: \[\underline{x} \cdot y = (ky) \cdot y = k(y \cdot y)\]
03

Calculate the magnitudes of the vectors

We need to calculate the magnitudes of the vectors \(\underline{x}\) and \(y\) as follows: \[|\underline{x}| = |ky|\] \[|y| = |y|\]
04

Find the product of the magnitudes

Now we can find the product of the magnitudes of the two vectors by multiplying their magnitudes: \[|\underline{x}||y| = |ky||y|\]
05

Compare the dot product and product of magnitudes

We can compare the dot product of \(\underline{x}\) and \(y\) from Step 2 and the product of their magnitudes from Step 4: \[\underline{x} \cdot y = k(y \cdot y)\] \[|\underline{x}||y| = |ky||y|\] Since \(k < 0\), the dot product of \(\underline{x}\) and \(y\) will be negative, while the product of the magnitudes will be positive. Therefore, the statement: \[\underline{x} \cdot y < |\underline{x}||y|\] is the correct option, which corresponds to the answer (d).

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Most popular questions from this chapter

\((1 / \mathrm{c}),(1 / \mathrm{c}),(1 / \mathrm{c})\) 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) \(\pm \sqrt{3}\)

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

\(\underline{\mathrm{a}}\) and \(\underline{\mathrm{b}}\) are unit vector. If \(\left(\underline{\mathrm{a}}^{\wedge} \underline{\mathrm{b}}\right)=\theta\) and \(|\underline{\mathrm{a}}-\underline{\mathrm{b}}|<1\) then \(\theta \in\) (a) \([0,(\pi / 3)]\) (b) \([(2 \pi / 3) \cdot(4 \pi / 3)]\) (c) \([(\pi / 8) \cdot(\pi / 2)]\) (d) \([0,(\pi / 3)]\)

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