Question

1 Find the angle between A=2x+2j−kand B=6i−3j+2k
2 Determine the value of a so that A=2i+aj+kand B=4i−2j−2k are perpendicular
3 Determine a unit vector perpendicular to the plane of A=2i−6j−3k and B=4i+3j−k
4 Find the work done in moving an object along a vector r=3i+2j−5k
5 Given that A=2i−j+3kand B=3i+2j−k, find A⋅B

Answer 1

Answer on Question #59203 – Math – Linear Algebra

If $\mathbf{u}$ and $\mathbf{v}$ are nonzero vectors (in $R^2$ or $R^3$) and if $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$, then the dot product of $\mathbf{u}$ and $\mathbf{v}$ is denoted by $u \cdot v$ and is defined as

$$u \cdot v = |u| \cdot |v| \cdot \cos \theta$$

or in terms of coordinates as

$$u \cdot v = u_x v_x + u_y v_y + u_z v_z,$$

where $u = (u_x; u_y; u_z)$, $v = (v_x; v_y; v_z)$.

Besides,

$u \cdot v = 0 \Leftrightarrow u$ and $v$ are perpendicular.

Question

1. Find the angle between $A = 2i + 2j - k$ and $B = 6i - 3j + 2k$.

Solution

$$\cos \theta = \frac{A \cdot B}{|A \parallel B|};$$

- $A = 2i + 2j - k = (2; 2; -1)$;

- $B = 6i - 3j + 2k = (6; -3; 2)$;

- $A \cdot B = 2 \cdot 6 + 2 \cdot (-3) + (-1) \cdot 2 = 12 - 6 - 2 = 4$;

- $|A| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{9} = 3$;

- $|B| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{49} = 7$;

- $\cos \theta = \frac{4}{3 \cdot 7} = \frac{4}{21}$;

- $\theta = \arccos \frac{4}{21}$, hence $\theta \approx 79.02{}^\circ$.

Answer

$\theta = \arccos \frac{4}{21}$, $\theta \approx 79.02{}^\circ$.

Question

2. Determine the value of $a$ so that $A = 2i + aj + k$ and $B = 4i - 2j - 2k$ are perpendicular.

Solution

- $A = 2i + aj + k = (2; a; 1)$

- $B = 4i - 2j - 2k = (4; -2; -2)$

- $A \cdot B = 2 \cdot 4 + a \cdot (-2) + 1 \cdot (-2) = 8 - 2a - 2 = 6 - 2a$

- $A \cdot B = 0$, then $6 - 2a = 0; 2a = 6; a = 3$.

Answer

$a = 3$.

Question

3. Determine a unit vector perpendicular to the plane of $A = 2i - 6j - 3k$ and $B = 4i + 3j - k$

Solution

A=2i-6j-3k=(2; -6; -3)

B=4i+3j-k=(4; 3; -1)

Let u be a unit vector perpendicular to the plane of A=2i-6j-3k and B=4i+3j-k, then

$$u = \frac {A \times B}{| A \times B |};$$ $$\begin{array}{l} A \times B = \left| \begin{array}{ccc} i & j & k \\ 2 & -6 & -3 \\ 4 & 3 & -1 \end{array} \right| = i \left| \begin{array}{cc} -6 & -3 \\ 3 & -1 \end{array} \right| - j \left| \begin{array}{cc} 2 & -3 \\ 4 & -1 \end{array} \right| + k \left| \begin{array}{cc} 2 & -6 \\ 4 & 3 \end{array} \right| = \\ = i (6 + 9) - j (- 2 + 12) + k (6 + 24) = 15i - 10j + 30k \\ \end{array}$$ $$\begin{array}{l} \left| A \times B \right| = \sqrt {15 ^ {2} + (- 10) ^ {2} + 30 ^ {2}} = \sqrt {225 + 100 + 900} = \sqrt {1225} = 35; \\ u = \frac {A \times B}{| A \times B |} = \frac {1}{35} (15i - 10j + 22k) = \frac {15}{35} i - \frac {10}{35} j + \frac {30}{35} k = \frac {3}{7} i - \frac {2}{7} j + \frac {6}{7} k; \\ \end{array}$$

Answer:

$$u = \frac {3}{7} i - \frac {2}{7} j + \frac {6}{7} k.$$

Question

4 Find the work done in moving an object along a vector $r = 3i + 2j - 5k$.

Solution

The object is moved by the force, and

the work down in moving an object along a vector $r$, if the applied force $F$, is defined by the dot product

$$A = F \cdot r$$

Let the vector of applied force be $F = x_i + y_j + z_k = (x, y, z)$,

then $A = F \cdot r = 3x + 2y - 5z$.

Answer: $A = F \cdot r = 3x + 2y - 5z$ for $F = x_i + y_j + z_k$.

Question

5 Given that $A = 2i - j + 3k$ and $B = 3i + 2j - k$, find $A \cdot B$

Solution

A=2i-j+3k=(2;-1;3)

B=3i+2j-k=(3; 2; -1)

A·B=2*3+(-1)*2+3*(-1)=6-2-3=1;

Answer: $A \cdot B = 1$.

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