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1. ABCDA B C D is a parallelogram, and $|\overrightarrow{P D}|=|\overrightarrow{D ...
Mar 26, 2024
1. ABCDA B C D is a parallelogram, and ∣PD→∣=∣DA→∣|\overrightarrow{P D}|=|\overrightarrow{D A}|. a. Determine which vectors (if any) are equal to AB→,BA→,AD→,CB→\overrightarrow{A B}, \overrightarrow{B A}, \overrightarrow{A D}, \overrightarrow{C B}, and AP→\overrightarrow{A P}. b. Explain why ∣PD→∣=∣BC→∣|\overrightarrow{P D}|=|\overrightarrow{B C}|.
Solution by Steps
step 1
Identify the given vectors in the parallelogram ABCD
step 2
In a parallelogram, opposite sides are equal and parallel, so AB→=CD→\overrightarrow{AB} = \overrightarrow{CD} and AD→=BC→\overrightarrow{AD} = \overrightarrow{BC}
step 3
Since vectors have both magnitude and direction, BA→\overrightarrow{BA} is the same magnitude as AB→\overrightarrow{AB} but in the opposite direction, so BA→=−AB→\overrightarrow{BA} = -\overrightarrow{AB}
step 4
Similarly, CB→\overrightarrow{CB} is the same magnitude as BC→\overrightarrow{BC} but in the opposite direction, so CB→=−BC→\overrightarrow{CB} = -\overrightarrow{BC}
step 5
Given that ∣PD→∣=∣DA→∣|\overrightarrow{PD}| = |\overrightarrow{DA}|, and assuming they have the same direction, PD→=DA→\overrightarrow{PD} = \overrightarrow{DA}
step 6
To find AP→\overrightarrow{AP}, we can use vector addition: AP→=AD→+DP→\overrightarrow{AP} = \overrightarrow{AD} + \overrightarrow{DP}. Since DP→=−PD→\overrightarrow{DP} = -\overrightarrow{PD}, we have AP→=AD→−DA→=0→\overrightarrow{AP} = \overrightarrow{AD} - \overrightarrow{DA} = \overrightarrow{0}
Answer
AB→=CD→,AD→=BC→,BA→=−AB→,CB→=−BC→,AP→=0→\overrightarrow{AB} = \overrightarrow{CD}, \overrightarrow{AD} = \overrightarrow{BC}, \overrightarrow{BA} = -\overrightarrow{AB}, \overrightarrow{CB} = -\overrightarrow{BC}, \overrightarrow{AP} = \overrightarrow{0}
Key Concept
Vector properties in a parallelogram
Explanation
In a parallelogram, opposite sides are equal and parallel, and the direction of a vector is reversed when the order of its points is reversed. The zero vector results from adding a vector to its own negative.
step 1
Understand that in a parallelogram, opposite sides are congruent
step 2
Since ∣PD→∣=∣DA→∣|\overrightarrow{PD}| = |\overrightarrow{DA}| and DA→=BC→\overrightarrow{DA} = \overrightarrow{BC} by the properties of a parallelogram, it follows that ∣PD→∣=∣BC→∣|\overrightarrow{PD}| = |\overrightarrow{BC}|
Answer
∣PD→∣=∣BC→∣|\overrightarrow{PD}| = |\overrightarrow{BC}|
Key Concept
Congruent sides in a parallelogram
Explanation
In a parallelogram, opposite sides are congruent, which means they have the same length. Therefore, the vectors representing these sides have the same magnitude.
2. The diagram below represents a rectangular prism. State a single vector equal to each of the following. a. RQ→+RS→\overrightarrow{R Q}+\overrightarrow{R S} c. PW→+WS→\overrightarrow{P W}+\overrightarrow{W S} e. PW→−VP→\overrightarrow{P W}-\overrightarrow{V P} b. RQ→+QV→\overrightarrow{R Q}+\overrightarrow{Q V} d. (RQ→+RS→)+VU→(\overrightarrow{R Q}+\overrightarrow{R S})+\overrightarrow{V U} f. PW→+WR→+RQ→\overrightarrow{P W}+\overrightarrow{W R}+\overrightarrow{R Q}
Solution by Steps
step 2
Since RQ→\overrightarrow{RQ} and RS→\overrightarrow{RS} share the same initial point, we can add them head-to-tail
step 3
The resulting vector goes from the tail of RQ→\overrightarrow{RQ} to the head of RS→\overrightarrow{RS}, which is QS→\overrightarrow{QS}
A
Key Concept
Vector Addition
Explanation
When two vectors share the same initial point, the resultant vector is the one that starts at the tail of the first and ends at the head of the second.
Solution by Steps
step 2
Since PW→\overrightarrow{PW} and WS→\overrightarrow{WS} are consecutive vectors, we can add them head-to-tail
step 3
The resulting vector goes from the tail of PW→\overrightarrow{PW} to the head of WS→\overrightarrow{WS}, which is PS→\overrightarrow{PS}
C
Key Concept
Vector Addition
Explanation
When adding consecutive vectors, the resultant vector is the one that starts at the tail of the first and ends at the head of the second.
Solution by Steps
step 2
Since VP→\overrightarrow{VP} is the reverse of PV→\overrightarrow{PV}, subtracting it is the same as adding PV→\overrightarrow{PV}
step 3
The resulting vector goes from the tail of PW→\overrightarrow{PW} to the head of PV→\overrightarrow{PV}, which is WV→\overrightarrow{WV}
E
Key Concept
Vector Subtraction
Explanation
Subtracting a vector is the same as adding its reverse direction vector.
Solution by Steps
step 2
Since RQ→\overrightarrow{RQ} and QV→\overrightarrow{QV} are consecutive vectors, we can add them head-to-tail
step 3
The resulting vector goes from the tail of RQ→\overrightarrow{RQ} to the head of QV→\overrightarrow{QV}, which is RV→\overrightarrow{RV}
B
Key Concept
Vector Addition
Explanation
When adding consecutive vectors, the resultant vector is the one that starts at the tail of the first and ends at the head of the second.
Solution by Steps
step 2
From previous steps, we know that RQ→+RS→=QS→\overrightarrow{RQ}+\overrightarrow{RS} = \overrightarrow{QS}
step 3
Now we add QS→\overrightarrow{QS} and VU→\overrightarrow{VU}, which do not share an initial or terminal point, so we cannot simplify this to a single vector without additional information
D
Key Concept
Vector Addition with Non-Consecutive Vectors
Explanation
When vectors do not share an initial or terminal point, additional information is needed to simplify to a single vector.
Solution by Steps
step 2
Since PW→\overrightarrow{PW}, WR→\overrightarrow{WR}, and RQ→\overrightarrow{RQ} are consecutive vectors, we can add them head-to-tail
step 3
The resulting vector goes from the tail of PW→\overrightarrow{PW} to the head of RQ→\overrightarrow{RQ}, which is PQ→\overrightarrow{PQ}
F
Key Concept
Vector Addition
Explanation
When adding consecutive vectors, the resultant vector is the one that starts at the tail of the first and ends at the head of the last.
3. Two vectors, a⃗\vec{a} and b⃗\vec{b}, have a common starting point with angle of 120∘120^{\circ} between them. The vectors are such that a⃗∣=3\vec{a} \mid=3 and ∣b⃗∣=4|\vec{b}|=4. a. Calculate ∣a⃗+b⃗∣|\vec{a}+\vec{b}|. b. Calculate the angle between a⃗\vec{a} and a⃗+b⃗\vec{a}+\vec{b}.
Solution by Steps
step 1
To find the magnitude of the sum of two vectors, we use the formula ∣a⃗+b⃗∣=∣a⃗∣2+∣b⃗∣2+2∣a⃗∣∣b⃗∣cos⁡(θ)|\vec{a}+\vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos(\theta)}, where θ\theta is the angle between the vectors
step 2
Given ∣a⃗∣=3|\vec{a}|=3, ∣b⃗∣=4|\vec{b}|=4, and θ=120∘\theta = 120^\circ, we substitute these values into the formula: ∣a⃗+b⃗∣=32+42+2⋅3⋅4⋅cos⁡(120∘)|\vec{a}+\vec{b}| = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \cos(120^\circ)}
step 3
Calculate the cosine of 120∘120^\circ: cos⁡(120∘)=−12\cos(120^\circ) = -\frac{1}{2}
step 4
Substitute the cosine value into the equation: ∣a⃗+b⃗∣=9+16−2⋅3⋅4⋅(−12)|\vec{a}+\vec{b}| = \sqrt{9 + 16 - 2 \cdot 3 \cdot 4 \cdot \left(-\frac{1}{2}\right)}
step 5
Simplify the equation: ∣a⃗+b⃗∣=9+16+12=37|\vec{a}+\vec{b}| = \sqrt{9 + 16 + 12} = \sqrt{37}
Answer
∣a⃗+b⃗∣=37|\vec{a}+\vec{b}| = \sqrt{37}
Key Concept
Magnitude of the sum of two vectors
Explanation
The magnitude of the sum of two vectors can be found using the formula that includes the magnitudes of the individual vectors and the cosine of the angle between them.
step 1
To find the angle between a⃗\vec{a} and a⃗+b⃗\vec{a}+\vec{b}, we use the dot product formula: a⃗⋅(a⃗+b⃗)=∣a⃗∣∣a⃗+b⃗∣cos⁡(ϕ)\vec{a} \cdot (\vec{a}+\vec{b}) = |\vec{a}| |\vec{a}+\vec{b}| \cos(\phi), where ϕ\phi is the angle between a⃗\vec{a} and a⃗+b⃗\vec{a}+\vec{b}
step 2
We know that a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2 and a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡(θ)\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta), with θ=120∘\theta = 120^\circ
step 3
Substitute the known values: 32+3⋅4⋅cos⁡(120∘)=9−6=33^2 + 3 \cdot 4 \cdot \cos(120^\circ) = 9 - 6 = 3
step 4
Now we have a⃗⋅(a⃗+b⃗)=3\vec{a} \cdot (\vec{a}+\vec{b}) = 3
step 5
Substitute the dot product and magnitudes into the formula: 3=3⋅37⋅cos⁡(ϕ)3 = 3 \cdot \sqrt{37} \cdot \cos(\phi)
step 6
Solve for cos⁡(ϕ)\cos(\phi): cos⁡(ϕ)=33⋅37=137\cos(\phi) = \frac{3}{3 \cdot \sqrt{37}} = \frac{1}{\sqrt{37}}
step 7
Find the angle ϕ\phi by taking the inverse cosine: ϕ=cos⁡−1(137)\phi = \cos^{-1}\left(\frac{1}{\sqrt{37}}\right)
Answer
ϕ=cos⁡−1(137)\phi = \cos^{-1}\left(\frac{1}{\sqrt{37}}\right)
Key Concept
Angle between two vectors
Explanation
The angle between two vectors can be found using the dot product formula and solving for the cosine of the angle, then finding the angle itself using the inverse cosine function.
4. Determine all possible values for tt if the length of the vector x⃗=ty⃗\vec{x}=t \vec{y} is 4y⃗∣4 \vec{y} \mid.
Solution by Steps
step 1
Given the equation ∣t⋅y∣=4⋅∣y∣ |t \cdot y| = 4 \cdot |y| , we want to solve for t t
step 2
We can divide both sides of the equation by ∣y∣ |y| , assuming y≠0 y \neq 0 , to get ∣t∣=4 |t| = 4
step 3
The absolute value equation ∣t∣=4 |t| = 4 has two solutions: t=4 t = 4 and t=−4 t = -4
Answer
t=4 t = 4 or t=−4 t = -4
Key Concept
Solving Absolute Value Equations
Explanation
The absolute value of a number is its distance from zero on the number line, which means it is always non-negative. Therefore, ∣t∣=4 |t| = 4 implies that t t can be either 4 4 or −4 -4 because both numbers are at a distance of 4 4 from zero.
5. PQRSP Q R S is a quadrilateral where A,B,CA, B, C, and DD are the midpoints of SP,PQS P, P Q, QRQ R, and RSR S, respectively. Prove, using vector methods, that ABCDA B C D is a parallelogram.
5. PQRSP Q R S is a quadrilateral where A,B,CA, B, C, and DD are the midpoints of SP,PQS P, P Q, QRQ R, and RSR S, respectively. Prove, using vector methods, that ABCDA B C D is a parallelogram.
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