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Chapter 1

Q. 1.9

Given the vectors

A = 8i + 4j − 2k lb

B = 2j + 6k ft

C = 3i − 2j + 4k ft

calculate the following: (1) A \cdot B; (2) the orthogonal component of B in the direction of C; (3) the angle between A and C; (4) A×B; (5) a unit vector λ that is perpendicular to both A and B; and (6) A × B \cdot C .

Step-by-Step

Verified Solution

Part 1
From Eq. (1.21), the dot product of A and B is
A \cdot B=A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}=8(0)+4(2)+(−2)(6)=−4\:lb \cdot ft
The negative sign indicates that the angle between A and B is greater than 90^\circ.

Part 2
Letting θ be the angle between B and C, we obtain from Eq. (1.23)

B\cosθ=B \cdot λ_{C}=B\cdot\frac{C}{C}=(2j+6k)\cdot\frac{3i−2j+4k}{\sqrt{3^{2}+(−2)^{2}+4^{2}}}=\frac{(0)(3)+(2)(−2)+(6)(4)}{\sqrt{29}}=3.71\:ft

Part 3
Letting α be the angle between A and C, we find from Eq. (1.22)

\cos\alpha =\lambda _{A}\cdot\lambda _{C}=\frac{A}{A}\cdot \frac{C}{C}=\frac{8i+4j−2k}{\sqrt{8^2+4^2+(−2)^2}}\cdot\frac{3i−2j+4k}{\sqrt{3^2+(−2)^2+4^2} }=\frac{(8)(3)+(4)(−2)+(−2)(4)}{\sqrt{84}\sqrt{29}}=0.16209

which yields
α=80.7^\circ

Part 4
Referring to Eq. (1.28), the cross product of A and B is

A×B=\left|\begin{matrix}i&j&k\\A_{x}&A_{y}&A_{z}\\B_{x} &B_{y}&B_{z}\end{matrix}\right|=\left|\begin{matrix}i&j&k\\8&4&-2\\0&2&6\end{matrix}\right|=i\left|\begin{matrix}4&-2\\2&6\end{matrix}\right|−j\left|\begin{matrix}8&-2\\0&6\end{matrix} \right|+k\left|\begin{matrix}8&4\\0&2\end{matrix}\right| =28i−48j+16k\:lb\cdot ft

Part 5
The cross product A×B is perpendicular to both A and B. Therefore, a unit vector in that direction is obtained by dividing A×B, which was evaluated above, by its magnitude

\frac{A\times B}{\left|A\times B\right|}=\frac{28i−48j+16k}{\sqrt{28^2+(−48)^2+16^2}}=0.484i−0.830j+0.277k

Because the negative of this vector is also a unit vector that is perpendicular to both A and B, we obtain
λ=±(0.484i−0.830j+0.277k)

Part 6
The scalar triple product A×B \cdot C is evaluated using Eq.(1.30).

A×B\cdot C=\left|\begin{matrix}A_{x}&A_{y}& A_{z}\\B_{x}&B_{y}&B_{z}\\C_{x}&C_{y}&C_{z}\end{matrix}\right|=\left|\begin{matrix}8&4&-2\\0&2&6\\3&-2&4\end{matrix}\right|=8\left|\begin{matrix}2&6\\-2&4\end{matrix}\right|−4\left|\begin{matrix}0&6\\3&4\end{matrix}\right|+(−2)\left|\begin{matrix}0&2\\3&-2\end{matrix}\right|=160+72+12=244\:lb\cdot ft^2