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Chapter 10 · Vector Algebra

Example 10: PQ→\overrightarrow{PQ} for P(2,3,0)P(2,3,0)⁠, Q(−1,−2,−4)Q(-1,-2,-4)

ExamplesVector joining two points · Section 10.5.2
The question

Find the vector joining the points P(2, 3, 0) and Q(−1-1⁠, −2-2⁠, −4-4⁠) directed from P to Q.

Exam answer

The vector is directed from P to Q, so P is the initial point and Q the terminal point. The required vector is PQ→\overrightarrow{PQ}⁠.

PQ→=OQ→−OP→=(−1−2)i^+(−2−3)j^+(−4−0)k^\begin{aligned} \overrightarrow{PQ} &= \overrightarrow{OQ} - \overrightarrow{OP} \\ &= (-1 - 2)\hat i + (-2 - 3)\hat j \\ &\quad + (-4 - 0)\hat k \end{aligned}

∴ PQ→=−3i^−5j^−4k^\therefore\ \overrightarrow{PQ} = -3\hat i - 5\hat j - 4\hat k⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The vector from a point P1(x1,y1,z1)P_1(x_1, y_1, z_1) to a point P2(x2,y2,z2)P_2(x_2, y_2, z_2) is "terminal minus initial":

P1P2→=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^.\begin{aligned} \overrightarrow{P_1P_2} &= (x_2 - x_1)\hat i + (y_2 - y_1)\hat j \\ &\quad + (z_2 - z_1)\hat k. \end{aligned}

It comes from the triangle law OP1→+P1P2→=OP2→\overrightarrow{OP_1} + \overrightarrow{P_1P_2} = \overrightarrow{OP_2}⁠. "Directed from P to Q" tells us which point is the start: P.

Given, to find, and the plan
Given
P(2, 3, 0) and Q(−1-1⁠, −2-2⁠, −4-4⁠).
To find
The vector joining P and Q, directed from P to Q, that is PQ→\overrightarrow{PQ}⁠.
Plan
(1) Identify the initial and terminal points. (2) Use PQ→=OQ→−OP→\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}⁠. (3) Subtract coordinate by coordinate. (4) Check: P moved by PQ→\overrightarrow{PQ} must land on Q.

Basics you need

1Position vectors

The position vector of a point P(x,y,z)P(x, y, z) is OP→=xi^+yj^+zk^\overrightarrow{OP} = x\hat i + y\hat j + z\hat k (Sections 10.2 and 10.5.1). Basics: three numbers fix a point

2Triangle law

For any points O, P, Q: OP→+PQ→=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}⁠, so PQ→=OQ→−OP→\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} (Section 10.4). Lesson: adding arrows, head to tail

3Vector joining two points

P1P2→=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^\overrightarrow{P_1P_2} = (x_2 - x_1)\hat i + (y_2 - y_1)\hat j + (z_2 - z_1)\hat k⁠, and its magnitude is the distance P1P2P_1P_2 given by the distance formula (Section 10.5.2). Lesson: components

Solution

  1. Initial and terminal points

    "Directed from P to Q": the vector starts at P and ends at Q. So P is the initial point and Q is the terminal point, and the required vector is PQ→\overrightarrow{PQ}⁠.

    The same two points give two different vectors, PQ→\overrightarrow{PQ} and QP→=−PQ→\overrightarrow{QP} = -\overrightarrow{PQ}⁠, with the same length and opposite directions. The phrase "directed from P to Q" decides which one is wanted.
    The initial point is where a vector starts, the terminal point where it ends (Section 10.2). Lesson: a vector is a length with a direction

  2. Write PQ→\overrightarrow{PQ} through the origin

    In the triangle OPQ, the triangle law gives OP→+PQ→=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}⁠, so

    PQ→=OQ→−OP→,\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP},

    with OP→=2i^+3j^+0k^\overrightarrow{OP} = 2\hat i + 3\hat j + 0\hat k and OQ→=−i^−2j^−4k^\overrightarrow{OQ} = -\hat i - 2\hat j - 4\hat k⁠.

    X Y Z O P(2, 3, 0) Q(−1, −2, −4) PQ OP OQ OP + PQ = OQ so PQ = OQ − OP
    P(2, 3, 0) lies in the xy-plane; Q(−1,−2,−4-1, -2, -4⁠) lies below it. Going from O to P and then from P to Q gives the same result as going straight from O to Q: OP→+PQ→=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}⁠. True (orthographic) view, one scale on all three axes; dashed lines show the coordinates.
    The points are given by coordinates, which are the components of their position vectors. The triangle law turns the unknown vector PQ→\overrightarrow{PQ} into a difference of two known position vectors.
    Triangle law (Section 10.4); P1P2→=OP2→−OP1→\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1} (Section 10.5.2, Fig 10.15). Lesson: adding arrows, head to tail

  3. Subtract coordinate by coordinate

    PQ→=(−1−2)i^+(−2−3)j^+(−4−0)k^=−3i^−5j^−4k^.\begin{aligned} \overrightarrow{PQ} &= (-1 - 2)\hat i + (-2 - 3)\hat j \\ &\quad + (-4 - 0)\hat k \\ &= -3\hat i - 5\hat j - 4\hat k. \end{aligned}

    So PQ→=−3i^−5j^−4k^\overrightarrow{PQ} = -3\hat i - 5\hat j - 4\hat k.

    The difference of two vectors in component form is found by subtracting corresponding components, and it is always terminal minus initial: Q's coordinates first, then P's. Each component is negative because each coordinate of Q is smaller than the same coordinate of P.
    a⃗−b⃗=(a1−b1)i^+(a2−b2)j^+(a3−b3)k^\vec a - \vec b = (a_1 - b_1)\hat i + (a_2 - b_2)\hat j + (a_3 - b_3)\hat k (Section 10.5.1 (ii)). Lesson: Examples 4–10

  4. Check by landing on Q

    Start at P and add the components of PQ→\overrightarrow{PQ}⁠:

    x: 2+(−3)=−1,y: 3+(−5)=−2,z: 0+(−4)=−4,\begin{aligned} x&:\ 2 + (-3) = -1, \\ y&:\ 3 + (-5) = -2, \\ z&:\ 0 + (-4) = -4, \end{aligned}

    which gives (−1,−2,−4)(-1, -2, -4)⁠, the point Q. The length is ∣PQ→∣=(−3)2+(−5)2+(−4)2=50=52|\overrightarrow{PQ}| = \sqrt{(-3)^2 + (-5)^2 + (-4)^2} = \sqrt{50} = 5\sqrt2⁠, the distance PQ.

    OP→+PQ→\overrightarrow{OP} + \overrightarrow{PQ} must equal OQ→\overrightarrow{OQ}⁠; adding back is the triangle law used in the other direction. The length agrees with the distance formula between P and Q, as Section 10.5.2 says it must.
    The magnitude of P1P2→\overrightarrow{P_1P_2} is the distance between P1P_1 and P2P_2 (Section 10.5.2). Basics: use Pythagoras twice
Result

PQ→=−3i^−5j^−4k^\overrightarrow{PQ} = -3\hat i - 5\hat j - 4\hat k⁠, of length 525\sqrt2⁠.

(Textbook answer: PQ→=(−1−2)i^+(−2−3)j^+(−4−0)k^\overrightarrow{PQ} = (-1-2)\hat i + (-2-3)\hat j + (-4-0)\hat k⁠, i.e. PQ→=−3i^−5j^−4k^\overrightarrow{PQ} = -3\hat i - 5\hat j - 4\hat k⁠.)

Check it with numbers

OP→+PQ→=(2−3)i^+(3−5)j^+(0−4)k^=−i^−2j^−4k^=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = (2 - 3)\hat i + (3 - 5)\hat j + (0 - 4)\hat k = -\hat i - 2\hat j - 4\hat k = \overrightarrow{OQ}⁠.

The length: 9+25+16=50≈7.071068\sqrt{9 + 25 + 16} = \sqrt{50} \approx 7.071068⁠, and 52≈5×1.414214=7.0710685\sqrt2 \approx 5 \times 1.414214 = 7.071068⁠. The reverse vector QP→=3i^+5j^+4k^\overrightarrow{QP} = 3\hat i + 5\hat j + 4\hat k has the same length and the opposite direction.

Common mistakes
  1. Initial minus terminal. OP→−OQ→=3i^+5j^+4k^\overrightarrow{OP} - \overrightarrow{OQ} = 3\hat i + 5\hat j + 4\hat k is QP→\overrightarrow{QP}⁠, directed from Q to P.
  2. Sign slips with negative coordinates. −1−2=−3-1 - 2 = -3⁠, −2−3=−5-2 - 3 = -5⁠, −4−0=−4-4 - 0 = -4⁠; bracket each subtraction.
  3. Adding the position vectors. OP→+OQ→=i^+j^−4k^\overrightarrow{OP} + \overrightarrow{OQ} = \hat i + \hat j - 4\hat k has nothing to do with the vector from P to Q.
What to take away
  1. PQ→=OQ→−OP→\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}⁠: terminal minus initial, coordinate by coordinate.
  2. "Directed from P to Q" makes P the initial point.
  3. Check by adding PQ→\overrightarrow{PQ} to P: you must land on Q.

Every fact used, and where it comes from

FactUsed inSource
Initial and terminal points; position vector of a pointSteps 1, 2Section 10.2
Triangle law OP→+PQ→=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}Steps 2, 4Section 10.4 (Lesson: adding arrows)
Difference of vectors by componentsStep 3Section 10.5.1 (ii)
Vector joining two points and its magnitudeSteps 2, 4Section 10.5.2 and Fig 10.15 (Lesson: components)
Negative of a vector: QP→=−PQ→\overrightarrow{QP} = -\overrightarrow{PQ}Step 1Section 10.3
The book's working and answerResultExample 10, Section 10.5.2 (Lesson: Examples 4–10)
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