Pre-Calculus - Evaluate geometric vectors

Determine the product: $2\cdot \left \langle -2,-4,5\right \rangle$

$\left \langle -4,-8,10\right \rangle$

$\left \langle 0,-2,7\right \rangle$

$\left \langle -\frac{2\sqrt5}{15},-\frac{4\sqrt5}{15}, \frac{\sqrt{5}}{3}\right \rangle$

Explanation

To find the product of the scalar and the vector, simply multiply the scalar throughout each term inside the vector. Do not confuse this with the dot product or the norm of a vector.

$2\cdot \left \langle -2,-4,5\right \rangle =\left \langle 2(-2),2(-4),2(5)\right \rangle= \left \langle -4,-8,10\right \rangle$

The answer is: $\left \langle -4,-8,10\right \rangle$

$\vec{A}=1.00i+2.00j$

$\vec{B}=3.00i-1.00j$

Find $\vec{A}+\vec{B}$ .

$\vec{R}=4.00i+1.00j$

$\vec{R}=3.00i+2.00j$

$\vec{R}=4.00i+2.00j$

$\vec{R}=3.16i+2.23j$

$\vec{R}=1.00i+4.00j$

Explanation

Finding the resultant requires us to add like components:

$\vec{R}=\vec{A}+\vec{B}$

$\vec{R}=(1.00i+2.00j)+(3.00i-1.00j)$

$\vec{R}=(1.00+3.00)i+(2.00-1.00)j$

$\vec{R}=4.00i+1.00j$

$\vec{A}=-i-2j$

What are the magnitude and angle, CCW from the x-axis, of $2\vec{A}$ ?

$m=4.47,\ \theta=243^o$

$m=-4.47,\ \theta=117^o$

$m=4.47,\ \theta=63.4^o$

$m=2.24,\ \theta=-63.4^o$

$m=4.47,\ \theta=127^o$

Explanation

When multiplying a vector by a constant (called scalar multiplication), we multiply each component by the constant.

$2\vec{A}=2(-1-2j)=-2i-4j$

The magnitude of this new vector is found with these new components:

$\sqrt{x^2+y^2}=\sqrt{(-2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=4.47$

To calculate the angle we must first find the inverse tangent of $\frac{y}{x}$ :

$\arctan(\frac{y}{x})=\arctan(\frac{-4}{-2})=\arctan(2)=63.4^o$

This is the principal arctan, but it is in the first quadrant while our vector is in the third. We to add the angle 180° to this value to arrive at our final answer.

$\theta=180^o+63.4^o=243^o$

Find the norm of the vector $\vec{v}=[2,4,3]$ .

$\left | \vec{v}\right |= \sqrt{29}\approx 5.385$

$\left | \vec{v}\right |= \sqrt{9} = 3$

$\left | \vec{v}\right |= 9$

$\left | \vec{v}\right |= 2\sqrt{6}\approx 4.899$

Explanation

We find the norm of a vector by finding the sum of each element squared and then taking the square root.

$\left | \vec{v} \right |= \sqrt{2^2+4^2+3^2}=\sqrt{29}$ .

Express a vector with magnitude 2.24 directed 63.4° CCW from the x-axis in unit vector form.

$\text{1.00i + 2.00j}$

$\text{2.00i + 1.00j}$

$\text{1.00i + 1.00j}$

$-2.00\text{i} - 1.00\text{j}$

$-1.00\text{i} - 2.00\text{j}$

Explanation

The x-coordinate is the magnitude times the cosine of the angle, while the y-coordinate is the magnitude times the sine of the angle.

$\text{magnitude}=2.24,\ \theta=63.4^o$

$x = 2.24 \cos(63.4^o) = 1.00$

$y = 2.24 \sin(63.4^o) = 2.00$

The resultant vector is: $\text{1.00i + 2.00j}$ .

Vector $\vec{A}$ has a magnitude of 3.61 and is at an angle of 124° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 2.24 at an anlge of 63.4° CCW from the x-axis.

Find $2\vec{A}-\vec{B}$ using the parallelogram graphical method.

$m=6.40,\ \theta=141^o$

$m=9.00,\ \theta=38.7^o$

$m=9.00,\ \theta=321^o$

$m=3.00,\ \theta=-38.7^o$

$m=6.40,\ \theta=33.7^o$

Explanation

First, construct the two vectors using ruler and protractor:

$2\vec{A}$ is twice the length of $\vec{A}$ , but in the same direction:

Since we are subtracting, reverse the direction of $\vec{B}$ :

Place the tails of $2\vec{A}$ and $-\vec{B}$ at the same point:

Construct a parallelogram:

Construct and measure the resultant using ruler and protractor:

$m_{\vec{R}}=6.40,\ \theta_{\vec{R}}=141^o$

$\vec{A}=-2.00i+3.00j$

$\vec{B}=-1.00i-2.00j$

Find the magnitude and angle CCW from the x-axis of $\vec{A}-\vec{B}$ using the nose-to-tail graphical method.

$m=5.10,\ \theta=101^o$

$m=4.00,\ \theta=101^o$

$m=5.10,\ \theta=258^o$

$m=4.90,\ \theta=78.7^o$

$m=-5.10,\ \theta=101^o$

Explanation

Construct $\vec{A}$ and $\vec{B}$ from their x- and y-components:

Since we are subtracting, reverse the direction of $\vec{B}$ :

Form $\vec{A}-\vec{B}=\vec{A}+(-\vec{B})$ by placing the tail of $-\vec{B}$ at the nose of $\vec{A}$ :

Construct and measure the resultant, $\vec{R}$ , from the tail of $\vec{A}$ to the nose of $-\vec{B}$ using a ruler and protractor:

$m_{\vec{R}}=5.10,\ \theta_{\vec{R}}=101^o$

Vector $\vec{A}$ has a magnitude of 3.61 and a direction 124° CCW from the x-axis. Express $2\vec{A}$ in unit vector form.

Explanation

For vector $2\vec{A}$ , the magnitude is doubled, but the direction remains the same.

For our calculation, we use a magnitude of:

$m=2\times3.61=7.22$

The x-coordinate is the magnitude times the cosine of the angle, while the y-coordinate is the magnitude times the sine of the angle.

$x = m \cos \theta = 7.22 \cos (124^o) = -4.00$

$y = m \sin \theta = 7.22 \sin (124^o) = 6.00$

The resultant vector is: .

Find the product of:

$3\cdot \left \langle -2,3\right \rangle$

$\left \langle -6,9\right \rangle$

$\left \langle6,9\right \rangle$

$\left \langle -15,15\right \rangle$

$\left \langle 15,15\right \rangle$

$\left \langle -1,6\right \rangle$

Explanation

When a scalar is multiplied to a vector, simply distribute that value for both terms in the vector.

$3\cdot \left \langle -2,3\right \rangle=\left \langle -6,9\right \rangle$

Find the product of the vector $\vec{r}= 11\hat{i}+7\hat{j}$ and the scalar .

$a\vec{r}=44 \hat{i}+28\vec{j}$

$a\vec{r}=72$

$a\vec{r}=44 \hat{j}+28\vec{i}$

$a\vec{r}=72 \hat{i}$

Explanation

When multiplying a vector by a scalar we multiply each component of the vector by the scalar and the result is a vector:

$a \vec{r}= 4 \cdot (11 \hat{i} + 7 \hat{j}) = 44 \hat{i}+28 \hat{j}$

Evaluate geometric vectors

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