Geometric Vectors

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Pre-Calculus › Geometric Vectors

Questions 1 - 10

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

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.

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

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

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.

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

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

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

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:

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

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

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

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

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.

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

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: $\text{1.00i + 2.00j}$ .

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: $\text{1.00i + 2.00j}$ .

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