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Chapter 6: Q5E (page 331)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

5. \(\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} \)

Short Answer

Expert verified

The value is \(\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{8}{{13}}}\\{\frac{{12}}{{13}}}\end{aligned}} \right)\).

Step by step solution

01

Inner product

Consider \({\mathop{\rm u}\nolimits} ,v,\) and \({\mathop{\rm w}\nolimits} \) as the vectors in \({\mathbb{R}^n}\) and consider \(c\) as the scalar. Then

  1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} \)
  2. \(\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) \cdot {\mathop{\rm w}\nolimits} = {\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} + {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm w}\nolimits} \)
  3. \(\left( {c{\mathop{\rm u}\nolimits} } \right) \cdot {\mathop{\rm v}\nolimits} = c\left( {{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} } \right) = {\mathop{\rm u}\nolimits} \cdot \left( {c{\mathop{\rm v}\nolimits} } \right)\)
  4. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \ge 0\)and \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 0\) if and only if \({\mathop{\rm u}\nolimits} = 0\).
02

Compute

\(\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} \)

It is given that \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\).

Evaluate \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} \) and \({\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\\ &= \left( { - 1} \right)\left( 4 \right) + 2\left( 6 \right)\\ &= - 4 + 12\\ &= 8\end{aligned}\)

\(\begin{aligned}{c}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\\ &= {\left( 4 \right)^2} + {\left( 6 \right)^2}\\ &= 16 + 36\\ &= 52\end{aligned}\)

Compute \(\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} \) as shown below:

\(\begin{aligned}{c}\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} &= \frac{8}{{52}}\left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\\ &= \frac{2}{{13}}\left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{\frac{8}{{13}}}\\{\frac{{12}}{{13}}}\end{aligned}} \right)\end{aligned}\)

Thus, the value is \(\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{8}{{13}}}\\{\frac{{12}}{{13}}}\end{aligned}} \right)\).

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Most popular questions from this chapter

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram鈥揝chmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

Find the distance between \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\2\end{aligned}} \right)\) and \({\mathop{\rm z}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\{ - 1}\\8\end{aligned}} \right)\).

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)

In Exercises 9-12, find a unit vector in the direction of the given vector.

10. \(\left( {\begin{aligned}{*{20}{c}}{ - 6}\\4\\{ - 3}\end{aligned}} \right)\)

In exercises 1-6, determine which sets of vectors are orthogonal.

  1. \(\left[ {\begin{array}{*{20}{c}}{ - 1}\\4\\{ - 3}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}5\\2\\1\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\{ - 4}\\{ - 7}\end{array}} \right]\)
See all solutions

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