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## 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) $A × B \cdot C$.

## 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).

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