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Chapter 10 · Vector Algebra

Miscellaneous · Q2: Scalar components and magnitude of PQ→\overrightarrow{PQ} for P(x1,y1,z1)P(x_1,y_1,z_1)⁠, Q(x2,y2,z2)Q(x_2,y_2,z_2)

Miscellaneous ExerciseVector joining two points · Section 10.5.2
The question

Find the scalar components and magnitude of the vector joining the points P(x1x_1⁠, y1y_1⁠, z1z_1⁠) and Q(x2x_2⁠, y2y_2⁠, z2z_2⁠).

Exam answer

The position vectors of P and Q are

OP→=x1i^+y1j^+z1k^,OQ→=x2i^+y2j^+z2k^\begin{aligned} \overrightarrow{OP} &= x_1\hat i + y_1\hat j + z_1\hat k, \\ \overrightarrow{OQ} &= x_2\hat i + y_2\hat j + z_2\hat k \end{aligned}

By the triangle law, OP→+PQ→=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}⁠, so

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

Its magnitude is

∣PQ→∣=(x2−x1)2+(y2−y1)2+(z2−z1)2\begin{aligned} |\overrightarrow{PQ}| = \sqrt{\begin{aligned}&(x_2 - x_1)^2 + (y_2 - y_1)^2 \\ &+ (z_2 - z_1)^2\end{aligned}} \end{aligned}

∴\therefore the scalar components of PQ→\overrightarrow{PQ} are x2−x1x_2 - x_1⁠, y2−y1y_2 - y_1⁠, z2−z1z_2 - z_1⁠, and its magnitude is ∣PQ→∣|\overrightarrow{PQ}| as found above.

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

The one idea this problem is built on

The vector from P to Q is end minus start: PQ→=OQ→−OP→\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}⁠. Its components are the three coordinate differences, and its length comes from Pythagoras applied twice, which is the distance formula.

Given, to find, and the plan
Given
The points P(x1x_1⁠, y1y_1⁠, z1z_1⁠) and Q(x2x_2⁠, y2y_2⁠, z2z_2⁠).
To find
The scalar components of PQ→\overrightarrow{PQ} and its magnitude ∣PQ→∣|\overrightarrow{PQ}|⁠.
Plan
(1) Write the position vectors of P and Q. (2) Use the triangle law to get PQ→\overrightarrow{PQ}⁠. (3) Read off the scalar components. (4) Apply the length formula. (5) Check on two sample points.

Basics you need

1Position vector

The point (x,y,z)(x, y, z) has position vector xi^+yj^+zk^x\hat i + y\hat j + z\hat k from the origin O (Section 10.5.1). Basics: three numbers fix a point

2Triangle law

Going O to P and then P to Q is the same displacement as going O to Q: OP→+PQ→=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ} (Section 10.4). Lesson: adding arrows, head to tail

3Scalar components and length

For r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠, the scalar components are xx⁠, yy⁠, zz and ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

Solution

  1. The position vectors of P and Q

    OP→=x1i^+y1j^+z1k^,OQ→=x2i^+y2j^+z2k^.\begin{aligned} \overrightarrow{OP} &= x_1\hat i + y_1\hat j + z_1\hat k, \\ \overrightarrow{OQ} &= x_2\hat i + y_2\hat j + z_2\hat k. \end{aligned}
    The coordinates of a point are exactly the scalar components of its position vector: to reach P from O you go x1x_1 along OX, y1y_1 along OY and z1z_1 along OZ.
    Position vector in component form (Section 10.5.1). Basics: three numbers fix a point

  2. The vector joining P to Q

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

    PQ→=OQ→−OP→=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^.\begin{aligned} \overrightarrow{PQ} &= \overrightarrow{OQ} - \overrightarrow{OP} \\ &= (x_2 - x_1)\hat i + (y_2 - y_1)\hat j \\ &\quad + (z_2 - z_1)\hat k. \end{aligned}
    Subtracting OP→\overrightarrow{OP} from both sides isolates PQ→\overrightarrow{PQ}⁠. Vectors in component form are subtracted component by component: the i^\hat i parts give x2−x1x_2 - x_1⁠, and so on. It is always "end minus start": Q's coordinates minus P's.
    Vector joining two points (Section 10.5.2); difference by components (Section 10.5.1, (ii)). Example 10 does the same with numbers. Lesson: components

  3. The scalar components

    The scalar components of PQ→\overrightarrow{PQ} are the coefficients of i^\hat i⁠, j^\hat j⁠, k^\hat k⁠:

    x2−x1,y2−y1,z2−z1.x_2 - x_1, \quad y_2 - y_1, \quad z_2 - z_1.
    "Scalar components" are the three numbers; the "vector components" would be (x2−x1)i^(x_2 - x_1)\hat i⁠, (y2−y1)j^(y_2 - y_1)\hat j and (z2−z1)k^(z_2 - z_1)\hat k⁠. The order matters: x1−x2x_1 - x_2 and so on are the components of QP→=−PQ→\overrightarrow{QP} = -\overrightarrow{PQ}⁠.
    Scalar and vector components (Section 10.5.1). Lesson: components

  4. The magnitude

    ∣PQ→∣=(x2−x1)2+(y2−y1)2+(z2−z1)2\begin{aligned} |\overrightarrow{PQ}| = \sqrt{\begin{aligned}&(x_2 - x_1)^2 + (y_2 - y_1)^2 \\ &+ (z_2 - z_1)^2\end{aligned}} \end{aligned}
    O X Y Z P Q x₂ − x₁ y₂ − y₁ z₂ − z₁ |PQ| = √(4 + 9 + 4) = √17
    A box with P and Q at opposite corners, drawn for the sample P(1, 1, 0), Q(3, 4, 2) in a true orthographic view (one scale for all three axes). Its edges are the scalar components x2−x1x_2 - x_1⁠, y2−y1y_2 - y_1⁠, z2−z1z_2 - z_1⁠, and the diagonal PQ is the magnitude.
    Apply ∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} with the three components from Step 3. Geometrically, the edges of the box in the figure are the differences, Pythagoras on the floor gives the floor diagonal, and Pythagoras again with the height gives PQ. This is the distance formula of coordinate geometry.
    Magnitude of the vector joining two points (Section 10.5.2). Basics: use Pythagoras twice

  5. Two quick checks

    Order does not matter for the length. Each square is unchanged when the difference is reversed: (x1−x2)2=(x2−x1)2(x_1 - x_2)^2 = (x_2 - x_1)^2⁠. So ∣QP→∣=∣PQ→∣|\overrightarrow{QP}| = |\overrightarrow{PQ}|⁠, as a distance should be.

    A sample. For P(1, 1, 0) and Q(3, 4, 2) (the figure), the components are 3−1=23 - 1 = 2⁠, 4−1=34 - 1 = 3⁠, 2−0=22 - 0 = 2 and ∣PQ→∣=4+9+4=17|\overrightarrow{PQ}| = \sqrt{4 + 9 + 4} = \sqrt{17}⁠.

    A general formula should give the right answer on a case we can work out directly, and should behave sensibly when the roles of P and Q are swapped. Both checks pass.
    QP→=−PQ→\overrightarrow{QP} = -\overrightarrow{PQ}⁠, the negative of a vector (Section 10.3). Lesson: six kinds of vectors
Result

PQ→=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^\overrightarrow{PQ} = (x_2 - x_1)\hat i + (y_2 - y_1)\hat j + (z_2 - z_1)\hat k⁠. Its scalar components are x2−x1x_2 - x_1⁠, y2−y1y_2 - y_1⁠, z2−z1z_2 - z_1⁠, and its magnitude is

∣PQ→∣=(x2−x1)2+(y2−y1)2+(z2−z1)2\begin{aligned} |\overrightarrow{PQ}| = \sqrt{\begin{aligned}&(x_2 - x_1)^2 + (y_2 - y_1)^2 \\ &+ (z_2 - z_1)^2\end{aligned}} \end{aligned}

(Textbook answer: x2−x1x_2 - x_1⁠, y2−y1y_2 - y_1⁠, z2−z1z_2 - z_1⁠; (x2−x1)2+(y2−y1)2+(z2−z1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}⁠.)

Check it with numbers

Sample P(1, 1, 0), Q(3, 4, 2): 17≈4.123106\sqrt{17} \approx 4.123106⁠. Pythagoras twice: the floor diagonal is 22+32=13≈3.605551\sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.605551⁠, then 13+22=17\sqrt{13 + 2^2} = \sqrt{17}⁠.

A second sample, P(2, 3, 0), Q(−1-1⁠, −2-2⁠, −4-4⁠): the components are −3-3⁠, −5-5⁠, −4-4 and the magnitude is 9+25+16=50=52≈7.071068\sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt2 \approx 7.071068⁠.

Common mistakes
  1. Start minus end. x1−x2x_1 - x_2⁠, y1−y2y_1 - y_2⁠, z1−z2z_1 - z_2 are the components of QP→\overrightarrow{QP}⁠, the opposite vector.
  2. Mixing up scalar and vector components. The scalar components are numbers; the vector components carry i^\hat i⁠, j^\hat j⁠, k^\hat k⁠.
  3. Square root of each term. a2+b2+c2\sqrt{a^2 + b^2 + c^2} is not ∣a∣+∣b∣+∣c∣|a| + |b| + |c|⁠; in the sample, 2+3+2=72 + 3 + 2 = 7⁠, not 17\sqrt{17}⁠.
What to take away
  1. PQ→\overrightarrow{PQ} = position vector of Q minus position vector of P.
  2. Its scalar components are the coordinate differences, end minus start.
  3. Its magnitude is the distance PQ: the square root of the sum of the squared differences.

Every fact used, and where it comes from

FactUsed inSource
Position vector xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠; scalar and vector componentsSteps 1, 3Section 10.5.1 (Basics: three numbers fix a point)
Triangle law of additionStep 2Section 10.4 (Lesson: adding arrows, head to tail)
Difference of vectors by componentsStep 2Section 10.5.1, (ii)
Vector joining two points and its magnitudeSteps 2, 4Section 10.5.2, Fig 10.15
∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Pythagoras twice)Step 4Section 10.5.1 (Basics: use Pythagoras twice)
Negative of a vectorStep 5Section 10.3
The answerResultNCERT answer key, Miscellaneous Exercise on Chapter 10, Q2 (Lesson: put it all together)
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