10 chioce items of dot products of vector

The dot product of two vectors $\mathbf{a} = \langle a_1, a_2, a_3 \rangle$ and $\mathbf{b} = \langle b_1, b_2, b_3 \rangle$ is given by $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$.

$\mathbf{a} \cdot \mathbf{b} = (2)(4) + (-3)(1) + (1)(-2)$
$\mathbf{a} \cdot \mathbf{b} = 8 - 3 - 2$
$\mathbf{a} \cdot \mathbf{b} = 5 - 2$
$\mathbf{a} \cdot \mathbf{b} = 3$

The correct answer is A) 3.

Question 2: 2. If the dot product of two non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ is zero ($\mathbf{u} \cdot \mathbf{v} = 0$), what can be concluded about the relationship between $\mathbf{u}$ and $\mathbf{v}$?
A) They are parallel.
B) They are perpendicular (orthogonal).
C) They are equal in magnitude.
D) They point in the same direction.

The dot product $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta$, where $\theta$ is the angle between the vectors.

If $\mathbf{u} \cdot \mathbf{v} = 0$ and $\mathbf{u}$ and $\mathbf{v}$ are non-zero vectors (meaning $|\mathbf{u}| \neq 0$ and $|\mathbf{v}| \neq 0$), then it must be that $\cos \theta = 0$.

For $\cos \theta = 0$, the angle $\theta$ must be $90^\circ$ or $\pi/2$ radians. This means the vectors are perpendicular.

The correct answer is B) They are perpendicular (orthogonal).

$\mathbf{p} \cdot \mathbf{q} = (1)(0) + (0)(1) = 0 + 0 = 0$.

$|\mathbf{p}| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1$.
$|\mathbf{q}| = \sqrt{0^2 + 1^2} = \sqrt{1} = 1$.

$\cos \theta = \frac{\mathbf{p} \cdot \mathbf{q}}{|\mathbf{p}| |\mathbf{q}|} = \frac{0}{(1)(1)} = 0$.

Since $\cos \theta = 0$, $\theta = 90^\circ$.

The correct answer is C) $90^\circ$.

The dot product of two vectors results in a scalar. Therefore, $\mathbf{b} \cdot \mathbf{c}$ is a scalar. You cannot take the dot product of a vector ($\mathbf{a}$) with a scalar ($\mathbf{b} \cdot \mathbf{c}$). This expression is ill-defined. So, A is false.

The commutative property states that the order of multiplication does not affect the result. For dot products, $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$ and $\mathbf{b} \cdot \mathbf{a} = b_1a_1 + b_2a_2 + b_3a_3$. Since scalar multiplication is commutative, these are equal. So, B is true.

$\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$. For any non-zero vector $\mathbf{a}$, $|\mathbf{a}|^2$ will be a positive number, not 0. It is 0 only if $\mathbf{a}$ is the zero vector. So, C is false for non-zero vectors.

This is an incorrect application of the distributive property. The distributive property states $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$. The given expression is not the distributive property. So, D is false.

The correct answer is B) $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ (Commutative property).

For a vector $\mathbf{v} = \langle v_1, v_2, v_3 \rangle$, its magnitude is $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$.

$\mathbf{v} \cdot \mathbf{v} = v_1v_1 + v_2v_2 + v_3v_3 = v_1^2 + v_2^2 + v_3^2$.

From Step 1 and Step 2, we see that $|\mathbf{v}|^2 = v_1^2 + v_2^2 + v_3^2 = \mathbf{v} \cdot \mathbf{v}$.
Therefore, $|\mathbf{v}| = \sqrt{\mathbf{v} \cdot \mathbf{v}}$.

The correct answer is B) $\sqrt{\mathbf{v} \cdot \mathbf{v}}$.

Two non-zero vectors are orthogonal if their dot product is zero. So, $\mathbf{u} \cdot \mathbf{v} = 0$.

$\mathbf{u} \cdot \mathbf{v} = (3)(2) + (k)(-4) + (-1)(6)$
$\mathbf{u} \cdot \mathbf{v} = 6 - 4k - 6$
$\mathbf{u} \cdot \mathbf{v} = -4k$

$-4k = 0$
$k = 0$

The correct answer is A) $k = 0$.

The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is given by $\text{comp}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$.

$\mathbf{a} \cdot \mathbf{b} = (1)(3) + (2)(4) = 3 + 8 = 11$.

$|\mathbf{b}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

$\text{comp}_{\mathbf{b}}\mathbf{a} = \frac{11}{5}$.

The correct answer is A) $\frac{11}{5}$.

The standard unit vectors are:
$\mathbf{i} = \langle 1, 0, 0 \rangle$
$\mathbf{j} = \langle 0, 1, 0 \rangle$
$\mathbf{k} = \langle 0, 0, 1 \rangle$

$\mathbf{i} \cdot \mathbf{j} = (1)(0) + (0)(1) + (0)(0) = 0 + 0 + 0 = 0$.

Since $\mathbf{i}$ and $\mathbf{j}$ are perpendicular (orthogonal) unit vectors, their dot product is 0.

The correct answer is B) 0.

$\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta$.

$|\mathbf{u}| = 3$
$|\mathbf{v}| = 4$
$\theta = 60^\circ$

$\cos 60^\circ = \frac{1}{2}$.

$\mathbf{u} \cdot \mathbf{v} = (3)(4)\left(\frac{1}{2}\right)$
$\mathbf{u} \cdot \mathbf{v} = 12 \cdot \frac{1}{2}$
$\mathbf{u} \cdot \mathbf{v} = 6$

The correct answer is A) 6.

$\mathbf{a} \cdot \mathbf{b} = (x)(3) + (2)(-1)$
$\mathbf{a} \cdot \mathbf{b} = 3x - 2$

$3x - 2 = 7$
$3x = 7 + 2$
$3x = 9$
$x = \frac{9}{3}$
$x = 3$

The correct answer is A) 3.