Sign up

Precalculus Flashcards: Finding Components Of Vectors

Study Finding Components Of Vectors in Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Precalculus

Finding Components Of Vectors

0 mastered0 still learning

0% Complete

QUESTION
1/ 36

State the formula for the component form of the vector from P(x1,y1)P(x_1,y_1) to Q(x2,y2)Q(x_2,y_2).

Tap card or press Space to flip

ANSWER

⟨x2−x1, y2−y1⟩\langle x_2-x_1,\,y_2-y_1\rangle. Terminal minus initial gives the displacement vector.

How well did you know it?

Card 1 / 36

What this deck covers

This deck focuses on Finding Components Of Vectors, giving you a quick way to review the definitions, rules, and examples that matter most for Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

All flashcards

Flashcard 1: State the formula for the component form of the vector from P(x1,y1)P(x_1,y_1) to Q(x2,y2)Q(x_2,y_2).

Answer: ⟨x2−x1, y2−y1⟩\langle x_2-x_1,\,y_2-y_1\rangle. Terminal minus initial gives the displacement vector.

Flashcard 2: What is the component form of the vector from P(−10,4)P(-10,4) to Q(−3,12)Q(-3,12)?

Answer: ⟨7, 8⟩\langle 7,\ 8\rangle. Calculate ⟨−3−(−10),12−4⟩=⟨7,8⟩\langle -3-(-10), 12-4\rangle = \langle 7, 8\rangle.

Flashcard 3: What operation gives vector components from initial point PP to terminal point QQ in the plane?

Answer: Subtract coordinates: Q−PQ-P. Terminal coordinates minus initial coordinates gives components.

Flashcard 4: What is the component form of the vector from P(−5,4)P(-5,4) to Q(−9,10)Q(-9,10)?

Answer: ⟨−4, 6⟩\langle -4,\,6\rangle. Calculate: ⟨−9−(−5),10−4⟩=⟨−4,6⟩\langle -9-(-5), 10-4\rangle = \langle -4, 6\rangle.

Flashcard 5: What is the component form of the vector from P(2,−1)P(2,-1) to Q(7,3)Q(7,3)?

Answer: ⟨5, 4⟩\langle 5,\,4\rangle. Calculate: ⟨7−2,3−(−1)⟩=⟨5,4⟩\langle 7-2, 3-(-1)\rangle = \langle 5, 4\rangle.

Flashcard 6: What is the component form of the vector from P(0,0)P(0,0) to Q(−3,5)Q(-3,5)?

Answer: ⟨−3, 5⟩\langle -3,\,5\rangle. Calculate: ⟨−3−0,5−0⟩=⟨−3,5⟩\langle -3-0, 5-0\rangle = \langle -3, 5\rangle.

Flashcard 7: What operation produces vector components from initial P(x1,y1)P(x_1,y_1) to terminal Q(x2,y2)Q(x_2,y_2)?

Answer: Subtract initial coordinates from terminal: Q−PQ-P. Vector components are found by terminal minus initial coordinates.

Flashcard 8: If PQ→=⟨a,b⟩\overrightarrow{PQ}=\langle a,b\rangle, what is QP→\overrightarrow{QP} in component form?

Answer: ⟨−a, −b⟩\langle -a,\ -b\rangle. Reversing direction negates both components.

Flashcard 9: Which vector is the opposite direction of PQ→\overrightarrow{PQ}: QP→\overrightarrow{QP} or PQ→\overrightarrow{PQ}?

Answer: QP→\overrightarrow{QP}. Reversing initial and terminal points reverses the vector.

Flashcard 10: Which option is the correct component form for AB→\overrightarrow{AB} if A(−1,2)A(-1,2) and B(3,−5)B(3,-5)?

Answer: ⟨4, −7⟩\langle 4,\,-7\rangle. Calculate: ⟨3−(−1),−5−2⟩=⟨4,−7⟩\langle 3-(-1), -5-2\rangle = \langle 4, -7\rangle.

Flashcard 11: Find and correct the error: A(1,4)A(1,4), B(6,0)B(6,0), claimed AB→=⟨−5,4⟩\overrightarrow{AB}=\langle -5,4\rangle.

Answer: Correct: AB→=⟨5, −4⟩\overrightarrow{AB}=\langle 5,\,-4\rangle. Should be ⟨6−1,0−4⟩=⟨5,−4⟩\langle 6-1, 0-4\rangle = \langle 5, -4\rangle, not negated.

Flashcard 12: What is the component form of the vector from P(5,5)P(5,5) to Q(5,−2)Q(5,-2)?

Answer: ⟨0, −7⟩\langle 0,\ -7\rangle. Calculate ⟨5−5,−2−5⟩=⟨0,−7⟩\langle 5-5, -2-5\rangle = \langle 0, -7\rangle.

Flashcard 13: What is QP→\overrightarrow{QP} in component form if PQ→=⟨7,−2⟩\overrightarrow{PQ}=\langle 7,-2\rangle?

Answer: ⟨−7, 2⟩\langle -7,\,2\rangle. Reverse vector has opposite components: negate each component.

Flashcard 14: What is the component form of the vector from P(12,−3)P\left(\frac{1}{2},-3\right) to Q(52,1)Q\left(\frac{5}{2},1\right)?

Answer: ⟨2, 4⟩\langle 2,\ 4\rangle. Calculate ⟨52−12,1−(−3)⟩=⟨2,4⟩\langle \frac{5}{2}-\frac{1}{2}, 1-(-3)\rangle = \langle 2, 4\rangle.

Flashcard 15: What is the component form of the vector from P(−6,0)P(-6,0) to Q(−1,−5)Q(-1,-5)?

Answer: ⟨5, −5⟩\langle 5,\ -5\rangle. Calculate ⟨−1−(−6),−5−0⟩=⟨5,−5⟩\langle -1-(-6), -5-0\rangle = \langle 5, -5\rangle.

Flashcard 16: Identify the error: a student wrote PQ→=⟨x1−x2, y1−y2⟩\overrightarrow{PQ}=\langle x_1-x_2,\ y_1-y_2\rangle for P→QP\to Q.

Answer: Correct is ⟨x2−x1, y2−y1⟩\langle x_2-x_1,\ y_2-y_1\rangle. Student reversed the subtraction order.

Flashcard 17: Find the initial point PP if Q(5,−4)Q(5,-4) and PQ→=⟨2,3⟩\overrightarrow{PQ}=\langle 2,3\rangle.

Answer: P(3,−7)P(3,-7). Subtract vector from terminal: (5,−4)−⟨2,3⟩=(3,−7)(5,-4) - \langle 2,3\rangle = (3,-7).

Flashcard 18: Find QQ if P(1,−2)P(1,-2) and PQ→=⟨4,5⟩\overrightarrow{PQ}=\langle 4,5\rangle.

Answer: Q(5,3)Q(5,3). Add vector components to initial point: Q=(1+4,−2+5)Q = (1+4, -2+5).

Flashcard 19: What is the component form of the vector from P(a,b)P(a,b) to Q(c,d)Q(c,d)?

Answer: ⟨c−a, d−b⟩\langle c-a,\,d-b\rangle. General formula: terminal minus initial coordinates.

Flashcard 20: Identify the component form of the vector from P(−3,1)P(-3,1) to Q(2,1)Q(2,1).

Answer: ⟨5, 0⟩\langle 5,\,0\rangle. Calculate: ⟨2−(−3),1−1⟩=⟨5,0⟩\langle 2-(-3), 1-1\rangle = \langle 5, 0\rangle.

Flashcard 21: Identify the component form of the vector from P(6,−3)P(6,-3) to Q(6,−9)Q(6,-9).

Answer: ⟨0, −6⟩\langle 0,\,-6\rangle. Calculate: ⟨6−6,−9−(−3)⟩=⟨0,−6⟩\langle 6-6, -9-(-3)\rangle = \langle 0, -6\rangle.

Flashcard 22: What is the component form of the vector from P(−2,−3)P(-2,-3) to Q(−2,4)Q(-2,4)?

Answer: ⟨0, 7⟩\langle 0,\ 7\rangle. Calculate ⟨−2−(−2),4−(−3)⟩=⟨0,7⟩\langle -2-(-2), 4-(-3)\rangle = \langle 0, 7\rangle.

Flashcard 23: What is the component form of the vector from P(9,−1)P(9,-1) to Q(2,−1)Q(2,-1)?

Answer: ⟨−7, 0⟩\langle -7,\ 0\rangle. Calculate ⟨2−9,−1−(−1)⟩=⟨−7,0⟩\langle 2-9, -1-(-1)\rangle = \langle -7, 0\rangle.

Flashcard 24: What is the component form of the vector from P(−2,−7)P(-2,-7) to Q(−2,1)Q(-2,1)?

Answer: ⟨0, 8⟩\langle 0,\,8\rangle. Calculate: ⟨−2−(−2),1−(−7)⟩=⟨0,8⟩\langle -2-(-2), 1-(-7)\rangle = \langle 0, 8\rangle.

Flashcard 25: What is the component form of the vector from P(0,0)P(0,0) to Q(−3,8)Q(-3,8)?

Answer: ⟨−3, 8⟩\langle -3,\ 8\rangle. Calculate ⟨−3−0,8−0⟩=⟨−3,8⟩\langle -3-0, 8-0\rangle = \langle -3, 8\rangle.

Flashcard 26: Find PP if Q(−1,6)Q(-1,6) and PQ→=⟨3,−2⟩\overrightarrow{PQ}=\langle 3,-2\rangle.

Answer: P(−4,8)P(-4,8). Subtract vector from terminal: P=(−1−3,6−(−2))P = (-1-3, 6-(-2)).

Flashcard 27: Find the terminal point QQ if P(2,3)P(2,3) and PQ→=⟨4,−1⟩\overrightarrow{PQ}=\langle 4,-1\rangle.

Answer: Q(6,2)Q(6,2). Add vector to initial point: (2,3)+⟨4,−1⟩=(6,2)(2,3) + \langle 4,-1\rangle = (6,2).

Flashcard 28: What is the component form of the vector from P(3,7)P(3,7) to Q(−4,1)Q(-4,1)?

Answer: ⟨−7, −6⟩\langle -7,\ -6\rangle. Calculate ⟨−4−3,1−7⟩=⟨−7,−6⟩\langle -4-3, 1-7\rangle = \langle -7, -6\rangle.

Flashcard 29: What is the component form of the vector from P(1.5,−2)P(1.5,-2) to Q(4.5,1)Q(4.5,1)?

Answer: ⟨3, 3⟩\langle 3,\,3\rangle. Calculate: ⟨4.5−1.5,1−(−2)⟩=⟨3,3⟩\langle 4.5-1.5, 1-(-2)\rangle = \langle 3, 3\rangle.

Flashcard 30: What is the component form of the vector from P(−4,6)P(-4,6) to Q(1,2)Q(1,2)?

Answer: ⟨5, −4⟩\langle 5,\ -4\rangle. Calculate ⟨1−(−4),2−6⟩=⟨5,−4⟩\langle 1-(-4), 2-6\rangle = \langle 5, -4\rangle.

Flashcard 31: Identify the xx-component of the vector from P(x1,y1)P(x_1,y_1) to Q(x2,y2)Q(x_2,y_2).

Answer: x2−x1x_2-x_1. The horizontal displacement from initial to terminal point.

Flashcard 32: Identify the yy-component of the vector from P(x1,y1)P(x_1,y_1) to Q(x2,y2)Q(x_2,y_2).

Answer: y2−y1y_2-y_1. The vertical displacement from initial to terminal point.

Flashcard 33: What is the component form of the vector from P(9,2)P(9,2) to Q(4,2)Q(4,2)?

Answer: ⟨−5, 0⟩\langle -5,\,0\rangle. Calculate: ⟨4−9,2−2⟩=⟨−5,0⟩\langle 4-9, 2-2\rangle = \langle -5, 0\rangle.

Flashcard 34: What is the component form of the vector from P(−4,6)P(-4,6) to Q(1,−2)Q(1,-2)?

Answer: ⟨5, −8⟩\langle 5,\,-8\rangle. Calculate: ⟨1−(−4),−2−6⟩=⟨5,−8⟩\langle 1-(-4), -2-6\rangle = \langle 5, -8\rangle.

Flashcard 35: What is the component form of the vector from P(−1.5,2)P(-1.5,2) to Q(0.5,−1)Q(0.5,-1)?

Answer: ⟨2, −3⟩\langle 2,\ -3\rangle. Calculate ⟨0.5−(−1.5),−1−2⟩=⟨2,−3⟩\langle 0.5-(-1.5), -1-2\rangle = \langle 2, -3\rangle.

Flashcard 36: What is the component form of the vector from P(3,8)P(3,8) to Q(−1,0)Q(-1,0)?

Answer: ⟨−4, −8⟩\langle -4,\,-8\rangle. Calculate: ⟨−1−3,0−8⟩=⟨−4,−8⟩\langle -1-3, 0-8\rangle = \langle -4, -8\rangle.