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Precalculus : Evaluate geometric vectors

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Example Questions

Example Question #1 : Geometric Vectors

Find the vector given by the product: $\frac{4}{7}<-3,5,14>$

Possible Answers:

$<\frac{-3}{7},\frac{5}{7},2>$

$<\frac{-12}{7},\frac{20}{7},8>$

<-12,20,56>

$<\frac{-21}{4},\frac{35}{4},\frac{49}{2}>$

<-21,35,98>

Correct answer:

$<\frac{-12}{7},\frac{20}{7},8>$

Explanation:

Given a scalar k and a vector v, the vector given by their products is defined component-wise:

k<v_1,v_2,...,v_n>=<kv_1,kv_2,...,kv_n> .

Here, our product is:

$\frac{4}{7}<-3,5,14>=<\frac{4}{7}(-3),\frac{4}{7}(5),\frac{4}{7}(14)>\\ \\ =<\frac{-12}{7},\frac{20}{7},8>$

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Example Question #1 : Geometric Vectors

This question refers to the previous question.

Simplify.

$4\cdot \left<3,6,9\right>$

Possible Answers:

$\left<12, 24, 6\right>$

$\left<12, 24, 36\right>$

$\left<12, 4, 6\right>$

$\left<2, 2, 3\right>$

$\left<1, 24, 3\right>$

Correct answer:

$\left<12, 24, 36\right>$

Explanation:

In order to simplify this problem we need to multiply the scalar factor to each component of the vector.

In our case the scalar factor is

Thus,

$4\cdot \left<3,6,9\right>\rightarrow \left<4\cdot 3, 4\cdot 6, 4\cdot 9\right>$

$\rightarrow \left<12, 24, 36\right>$

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Example Question #1 : Evaluate Geometric Vectors

Evaluate: $6\left \langle -2,-3,4\right \rangle$

Possible Answers:

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

$\left \langle 4,3,10 \right \rangle$

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

$\left \langle -12,-18,24 \right \rangle$

$\left \langle 12,18,-24 \right \rangle$

Correct answer:

$\left \langle -12,-18,24 \right \rangle$

Explanation:

In order to determine the final value of the vector, distribute the scalar among each term in the vector.

$6\left \langle -2,-3,4\right \rangle=\left \langle -12,-18,24 \right \rangle$

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Example Question #3 : Evaluate Geometric Vectors

If the vector from A(3,-3) to B(5.3) was multiplied by a scale factor of 3, what is the new vector $\underset{AB}{\rightarrow}$ ?

Possible Answers:

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

$\left \langle 2,6\right \rangle$

$\left \langle 6,0\right \rangle$

$\left \langle 6,18\right \rangle$

$\left \langle 24,18\right \rangle$

Correct answer:

$\left \langle 6,18\right \rangle$

Explanation:

To find $\underset{AB}{\rightarrow}$ :

Subtract vector A(3,-3) from B(5.3) .

$B-A=\left \langle 2,6\right \rangle$

Multiply this vector by a scale of 3.

$3\cdot\left \langle 2,6\right \rangle= \left \langle 6,18\right \rangle$

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Example Question #1 : Evaluate Geometric Vectors

Find the product of:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #1 : Geometric Vectors

When given a vector and a scalar what happens to the length and angle of when multiplied with ?

Possible Answers:

$\left | AV\right |=\left | V\right |$ or the length of the product is the same as the original vector.

The angle is unchanged

$\left | AV\right |=\left | V\right |$ or the length of the product is the same as the original vector.

The angle is multiplied by .

$\left | AV\right |=A\left | V\right |$ or the length of the product is times the length of the original vector.

The angle is multiplied by

$\left | AV\right |=A\left | V\right |$ or the length of the product is times the length of the original vector.

The angle is unchanged.

$\left | AV\right |=\sqrt{A}\left | V\right |$ or the length of the product is $\sqrt{A}$ times as long as the original vector.

The angle is multiplied by .

Correct answer:

$\left | AV\right |=A\left | V\right |$ or the length of the product is times the length of the original vector.

The angle is unchanged.

Explanation:

In simple terms is the hypotenuse of a triangle formed by the components of . So when you multiply by it mupltiplies all the componets by . This makes the length of the hypotensuse grow by as demonstrated by ($A$a)^2+($A$b)^2=($A$c)^2 from the Pythagorean Theorem.

For the same reasons the angle does not change because the new longer triangle will be a similar triangle to the original triangle.

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Example Question #2 : Geometric Vectors

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

Possible Answers:

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

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

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

Correct answer:

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

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

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Example Question #161 : Matrices And Vectors

Evaluate (2.3i + 7.4j - 4.6k) - (3.2i+5.1j - 2.4k)

Possible Answers:

None of the other answers

(1.2i+8.2j-1.7k)

(2.3i-0.9j-2.2k)

(-0.9i + 2.3j -2.2k)

(5.5i + 12.5j - 7.0k)

Correct answer:

(-0.9i + 2.3j -2.2k)

Explanation:

When adding two vectors, they need to be expanded into their components. Luckily, the problem statement gives us the vectors already in their component form. From here, we just need to remember that we can only add like components. So for this problem we get:

i: 2.3 - 3.2 = -0.9

j: 7.4-5.1 = 2.3

k: -4.6 + 2.4 = -2.2

Now we can combine those values to write out the complete vector:

-0.9i + 2.3j -2.2k

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Example Question #1 : Evaluate Geometric Vectors

What is the magnitude and angle for the following vector, measured CCW from the x-axis?

-2.00 i + 3.00 j

Possible Answers:

$m=2.24,\ \theta=-56.3^o$

$m=3.61,\ \theta=236^o$

$m=3.61,\ \theta=-124^o$

$m=1.00,\ \theta=124^o$

$m=3.61,\ \theta=124^o$

Correct answer:

$m=3.61,\ \theta=124^o$

Explanation:

The magnitude of the vector is found using the distance formula:

$\sqrt{x^2+y^2}=\sqrt{(-2)^2+(3)^2}=\sqrt{13}=3.61$

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

$\arctan(\frac{y}{x})=\arctan(\frac{3}{-2})=-56.3^o$

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

$\theta=180^o+(-56.3^o)=124^o$

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Example Question #2 : Evaluate Geometric Vectors

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.

Possible Answers:

4.00 i - 6.00 j

4.00 i + 6.00 j

-2.00 i + 3.00 j

6.00 i - 4.00 j

-4.00 i + 6.00 j

Correct answer:

-4.00 i + 6.00 j

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: -4.00 i + 6.00 j .

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Precalculus : Evaluate geometric vectors

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Geometric Vectors

Example Question #1 : Geometric Vectors

Example Question #1 : Evaluate Geometric Vectors

Example Question #3 : Evaluate Geometric Vectors

Example Question #1 : Evaluate Geometric Vectors

Example Question #1 : Geometric Vectors

Example Question #2 : Geometric Vectors

Example Question #161 : Matrices And Vectors

Example Question #1 : Evaluate Geometric Vectors

Example Question #2 : Evaluate Geometric Vectors

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