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Precalculus : Determine if Two Vectors Are Parallel or Perpendicular

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Example Question #31 : Determine If Two Vectors Are Parallel Or Perpendicular

Find the angle between the vectors $\small <4, -6>$ and $\small <3, 11>$ .

Possible Answers:

$\small 48.81^o$

$\small 89.54^o$

$\small 83.83^o$

$\small 131.05^o$

$\small 17.97 ^o$

Correct answer:

$\small 131.05^o$

Explanation:

To determine the angle between our two vectors, we can use the fact that for any 2 vectors $\small \vec{a}$ and $\small \vec{b}$ , $\small \vec{a} \cdot \vec{b} = ||\vec{a}||*||\vec{b}||*cos(\theta)$ where $\small ||\vec{a}||$ is the magnitude and $\small \theta$ is the angle between the 2 vectors, which is what we are looking for.

Working from the left, we can first find the dot product, $\small <4, -6> \cdot <3, 11> = 4*3 + -6*11 = 12 -66=-54$

Now we'll find the magnitudes of the two vectors by using the Pythagorean Theorem:

$\small 4^2 + (-6)^2 = C^2$

$\small 16 + 36 = C^2$

$\small 52 = C^2$ take the square root of both sides

$\small \sqrt{52}=C$

$\small 3^2 + 11^2 = C^2$

$\small 9+121 = C^2$

$\small 130 = C^2$

$\small \sqrt{130}=C$

Now we can plug these values back into the equation to start solving for theta:

$\small -54 = \sqrt{52}*\sqrt{130}*cos(\theta)$ multiply the two numbers inside the radicals:

$\small -54 = \sqrt {6,760}*cos(\theta)$ divide both sides by $\small \sqrt{6,760}$

$\small -0.6567 \approx cos(\theta)$ take the inverse cosine of both sides

$\small 131.05 \approx \theta$

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Example Question #542 : Pre Calculus

Find the angle between the vectors $\small \small (13, 20^o)$ and $\small <-2, -7>$ . Note that the first vector is in polar form and the second is in component form.

Possible Answers:

$\small \small 125.9 ^ o$

$\small 99.99^o$

$\small 85.4^o$

$\small 5.54 ^ o$

$\small 54.32^o$

Correct answer:

$\small \small 125.9 ^ o$

Explanation:

To find the angle between two vectors, we can use the fact that $\small \vec{a} \cdot \vec{b}= ||\vec{a}||*||\vec{b}||*cos(\theta)$ . In order to find the dot product, we need to convert the vector $\small (13, 20^o)$ to component form. This is easiest to do after drawing a quick sketch of the vector:

Vector 5

To find the vertical component, set up an equation involving sine, since the vertical component is the side of a right triangle across from the 20-degree angle:

$\small sin(20^o)=\frac{y}{13}$ evaluate the sine of 20, then multiply by 13

$\small 4.446 \approx y$

To find the horizontal component, set up an equation involving cosine, since the horizontal component is the side of a right triangle adjacent to the 20-degree angle:

$\small cos(20^o)= \frac{x}{13}$ evaluate the cosine of 20, then multiply by 13

$\small 12.216 \approx x$

The component form of this vector can be written as $\small <12.21, 4.44>$

Next, we can find the magnitude of the vector already in component form by using the Pythagorean Theorem:

$\small (-2)^2 + (-7)^2 = C^2$

$\small 4 + 49 = C^2$

$\small 53 = C^2$ take the square root of both sides

$\small \sqrt{53}=C$

Now we have all the information that we need to solve for theta.

$\small <12.21, 4.44> \cdot <-2, -7> = 13*\sqrt{53}*cos(\theta)$ Find the dot product for the left side:

$\small 12.21*-2 + 4.44*-7 = -55.5$ which we can now put back in the equation:

$\small -55.5 = 13*\sqrt{53}*cos(\theta)$ now divide both sides by $\small 13*\sqrt{53} \approx 94.64$

$\small 0.5864 \approx cos(\theta)$ take invese cosine of both sides

$\small \small 125.90^o \approx \theta$

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Example Question #33 : Matrices And Vectors

Find ${} \vec{a} \cdot \vec{b}$ where ${} \vec{a} = <3, 7>$ and $\small \small \vec{b} = (3\sqrt2, 45^o)$ . Note that $\small \vec{a}$ is in component form while $\small \vec{b}$ is in polar form.

Possible Answers:

$\small 315+9\sqrt2$

$\small 9+21\sqrt2$

$\small 30$

$\small 23$

$\small 21+9\sqrt2$

Correct answer:

$\small 30$

Explanation:

First, convert $\small \vec{b}$ to component form. The easiest way to do that would be with special right triangles, knowing that for a 45-45-90 triangle the ratio of the hypotenuse to both of the legs is $\small \sqrt2 : 1$ . So if the magnitude of the vector is $\small 3\sqrt2$ , then the components must both be 3.

This could also be found by setting up and solving the equations:

$\small cos(45)=\frac{x}{3\sqrt2}$ and $\small sin(45)=\frac{y}{3\sqrt2}$

Now that both vectors are component form, we see that we're multiplying

$\small <3,7> \cdot <3,3> = 3*3 + 7*3 = 9+21 = 30$

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Example Question #31 : Determine If Two Vectors Are Parallel Or Perpendicular

$\vec{v}=-4i+5j$

$\vec{w}=5i+4j$

Which of the following best explains whether the two vectors above are perpendicular or parallel?

Possible Answers:

Parallel, because their dot product is one.

Perpendicular, because their dot product is one.

Parallel, because their dot product is zero.

Neither perpendicular nor parallel, because their dot product is neither zero nor one.

Perpendicular, because their dot product is zero.

Correct answer:

Perpendicular, because their dot product is zero.

Explanation:

Two vectors are perpendicular if their dot product is zero, and parallel if their dot product is 1.

Take the dot product of our two vectors to find the answer:

$v_i\cdot w_i+v_j\cdot w_j=v \cdot w$

Using our given vectors:

$\vec{v}=-4i+5j$

$\vec{w}=5i+4j$

$v\cdot w=(-4\cdot 5)+(5\cdot 4)=-20+20=0$

Thus our two vectors are perpendicular.

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Precalculus : Determine if Two Vectors Are Parallel or Perpendicular

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #31 : Determine If Two Vectors Are Parallel Or Perpendicular

Example Question #542 : Pre Calculus

Example Question #33 : Matrices And Vectors

Example Question #31 : Determine If Two Vectors Are Parallel Or Perpendicular

All Precalculus Resources

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