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Chapter 17: Problem 1670

\(\underline{\mathrm{a}}, \underline{\mathrm{b}}\) and \(\underline{\text { c }}\) are non zero vectors, if \(\underline{\mathrm{a}}=8 \underline{\mathrm{b}}\) and \(\underline{\mathrm{c}}=-7 \underline{\mathrm{b}}\) then the angle between \(\underline{a}\) and \(\underline{c}\) is (a) \(\pi\) (b) 0 (c) \((\pi / 4)\) \((\mathrm{d})(\pi / 2)\)

Short Answer

Expert verified
(a) \(\pi\)

Step by step solution

01

Review the dot product formula

Recall the dot product formula for the angle \(\theta\) between two vectors \(\underline{\mathrm{a}}\) and \(\underline{\mathrm{c}}\): \[\cos \theta = \frac{\underline{\mathrm{a}} \cdot \underline{\mathrm{c}}}{\left\lVert \underline{\mathrm{a}} \right\rVert \times \left\lVert \underline{\mathrm{c}} \right\rVert}\]
02

Find the dot product of \(\underline{\mathrm{a}}\) and \(\underline{\mathrm{c}}\)

Since \(\underline{\mathrm{a}}=8 \underline{\mathrm{b}}\) and \(\underline{\mathrm{c}}=-7 \underline{\mathrm{b}}\), we can find their dot product as: \[\underline{\mathrm{a}} \cdot \underline{\mathrm{c}} = (8 \underline{\mathrm{b}}) \cdot (-7 \underline{\mathrm{b}}) = 8(-7) (\underline{\mathrm{b}} \cdot \underline{\mathrm{b}}) = -56 \left\lVert \underline{\mathrm{b}} \right\rVert^2\]
03

Find the magnitudes of \(\underline{\mathrm{a}}\) and \(\underline{\mathrm{c}}\)

Now, find the magnitudes of \(\underline{\mathrm{a}}\) and \(\underline{\mathrm{c}}\): \[\left\lVert \underline{\mathrm{a}} \right\rVert = \left\lVert 8 \underline{\mathrm{b}} \right\rVert = 8 \left\lVert \underline{\mathrm{b}} \right\rVert\] \[\left\lVert \underline{\mathrm{c}} \right\rVert = \left\lVert -7 \underline{\mathrm{b}} \right\rVert = 7 \left\lVert \underline{\mathrm{b}} \right\rVert\]
04

Calculate \(\cos \theta\)

Using the dot product formula and the results from Steps 2 and 3, we can find \(\cos \theta\): \[\cos \theta = \frac{-56 \left\lVert \underline{\mathrm{b}} \right\rVert^2}{(8 \left\lVert \underline{\mathrm{b}} \right\rVert)(7 \left\lVert \underline{\mathrm{b}} \right\rVert)} = \frac{-56}{(8)(7)} = -1\]
05

Determine the angle \(\theta\)

As \(\cos \theta = -1\), the angle \(\theta\) between \(\underline{\mathrm{a}}\) and \(\underline{\mathrm{c}}\) must be \(\pi\). Therefore, the correct option is: (a) \(\pi\)

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Most popular questions from this chapter

find the number of vectors in \(\mathrm{R}^{3}\) such the angle between X-axis and vectors are \((\pi / 3)\) (a) 1 (b) 2 (c) 4 (d) infinite times

\(\underline{\mathrm{a}}\) 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) \([0,(\pi / 3)]\)

if \(\underline{a}\) and \(\underline{b}\) are vector and \(\underline{a} \cdot \underline{b}<0,|\underline{a} \cdot \underline{b}|=|\underline{a} \times \underline{b}|\) then angle between \(\underline{a}\) and \(\underline{b}\) is = (a) \(\pi\) (b) \((7 \pi / 4)\) (c) \((\pi / 4)\) (d) \((3 \pi / 4)\)

\(\underline{a}=(2,-3,6)\) and \(\underline{b}=(-2,2,-1) .\) if \(\lambda=\) projection of a over \(\underline{b} /\) projection of \(\underline{b}\) over \(\underline{a}\). then \(\lambda=\) (a) \(\overline{(3 / 7)}\) (b) 7 (c) 3 (d) \((7 / 3)\)

\(\underline{\mathrm{a}} \cdot(2 \underline{\mathrm{b}}+2 \mathrm{c}) \times(3 \underline{\mathrm{a}}+3 \underline{\mathrm{b}}+3 \underline{\mathrm{c}})=\) (a) \([\underline{\mathrm{a}} \underline{\mathrm{b}} \underline{\mathrm{c}}]\) (b) \(3[\underline{\mathrm{a}} \underline{\mathrm{b}} \underline{\mathrm{c}}]\) (c) \(6[\underline{\mathrm{a}} \underline{\mathrm{b}} \underline{\mathrm{c}}]\) (d) 0
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