Class 12 Maths, Worked in Full Past papersGet updates
Chapter 10 · Vector Algebra

Exercise 10.2 · Q8: Unit vector along PQ→\overrightarrow{PQ}⁠, P(1,2,3)P(1,2,3)⁠, Q(4,5,6)Q(4,5,6)

Exercise 10.2Vector joining two points, unit vector · Sections 10.5 and 10.5.2
The question

Find the unit vector in the direction of vector PQ→\overrightarrow{PQ}⁠, where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

Exam answer

Here P is (1, 2, 3) and Q is (4, 5, 6).

PQ→=(4−1)i^+(5−2)j^+(6−3)k^=3i^+3j^+3k^\begin{aligned} \overrightarrow{PQ} &= (4 - 1)\hat i + (5 - 2)\hat j \\ &\quad + (6 - 3)\hat k \\ &= 3\hat i + 3\hat j + 3\hat k \end{aligned} ∣PQ→∣=9+9+9=27=33\begin{aligned} |\overrightarrow{PQ}| &= \sqrt{9 + 9 + 9} \\ &= \sqrt{27} = 3\sqrt3 \end{aligned}

The unit vector in the direction of PQ→\overrightarrow{PQ} is

PQ→∣PQ→∣=3i^+3j^+3k^33=13(i^+j^+k^)\begin{aligned} \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|} &= \frac{3\hat i + 3\hat j + 3\hat k}{3\sqrt3} \\ &= \frac{1}{\sqrt3}(\hat i + \hat j + \hat k) \end{aligned}

∴\therefore the unit vector is 13i^+13j^+13k^\dfrac{1}{\sqrt3}\hat i + \dfrac{1}{\sqrt3}\hat j + \dfrac{1}{\sqrt3}\hat k⁠.

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

The one idea this problem is built on

First turn the two points into one vector: PQ→\overrightarrow{PQ} is "terminal point minus initial point", here Q minus P. Then it is the usual unit-vector question: divide PQ→\overrightarrow{PQ} by its own length.

Given, to find, and the plan
Given
The points P(1, 2, 3) and Q(4, 5, 6).
To find
The unit vector in the direction of PQ→\overrightarrow{PQ}⁠, the direction from P to Q.
Plan
(1) Write PQ→\overrightarrow{PQ} in components. (2) Find ∣PQ→∣|\overrightarrow{PQ}|⁠. (3) Divide and simplify. (4) Check the length and the direction.

Basics you need

1Vector joining two points

For P1(x1,y1,z1)P_1(x_1, y_1, z_1) and P2(x2,y2,z2)P_2(x_2, y_2, z_2)⁠, P1P2→=(x2−x1)i^\overrightarrow{P_1P_2} = (x_2 - x_1)\hat i +(y2−y1)j^+(z2−z1)k^+ (y_2 - y_1)\hat j + (z_2 - z_1)\hat k (Section 10.5.2). Basics: three numbers fix a point

2Magnitude from components

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

3The unit vector

For a⃗≠0⃗\vec a \ne \vec 0⁠, a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a is the unit vector in the direction of a⃗\vec a (Section 10.5). Lesson: components

Solution

  1. Write PQ→\overrightarrow{PQ} in components

    PQ→=(4−1)i^+(5−2)j^+(6−3)k^=3i^+3j^+3k^.\begin{aligned} \overrightarrow{PQ} &= (4 - 1)\hat i + (5 - 2)\hat j + (6 - 3)\hat k \\ &= 3\hat i + 3\hat j + 3\hat k. \end{aligned}
    PQ→\overrightarrow{PQ} starts at P and ends at Q, so each component is "Q's coordinate minus P's coordinate". By the triangle law, OP→+PQ→=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}⁠, so PQ→=OQ→−OP→\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}⁠. Each coordinate of Q is 3 more than the matching coordinate of P, so every component is 3.
    P1P2→=OP2→−OP1→\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1} (Section 10.5.2, Fig 10.15). Basics: three numbers fix a point

  2. Find its length

    ∣PQ→∣=32+32+32=27=33.\begin{aligned} |\overrightarrow{PQ}| &= \sqrt{3^2 + 3^2 + 3^2} \\ &= \sqrt{27} = 3\sqrt3. \end{aligned}
    The length of PQ→\overrightarrow{PQ} is the distance PQ. Since 27=9×327 = 9 \times 3⁠, 27=33\sqrt{27} = 3\sqrt3⁠. The length is not 0, so P and Q are different points and the direction from P to Q makes sense.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. Divide by the length

    PQ→∣PQ→∣=3i^+3j^+3k^33=13(i^+j^+k^)=13i^+13j^+13k^.\begin{aligned} \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|} &= \frac{3\hat i + 3\hat j + 3\hat k}{3\sqrt3} \\ &= \frac{1}{\sqrt3}(\hat i + \hat j + \hat k) \\ &= \frac{1}{\sqrt3}\hat i + \frac{1}{\sqrt3}\hat j + \frac{1}{\sqrt3}\hat k. \end{aligned}

    This is the unit vector in the direction of PQ→\overrightarrow{PQ}.

    Dividing by the positive number 333\sqrt3 keeps the direction and makes the length 1. The common factor 3 cancels: 333=13\frac{3}{3\sqrt3} = \frac{1}{\sqrt3}⁠. Rationalised, each component is 33\frac{\sqrt3}{3}⁠.
    a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a (Section 10.5); λa⃗\lambda\vec a by components (Section 10.5.1, (iv)). Lesson: components

  4. Check the length and the direction

    (13)2×3=13×3=1.\left(\tfrac{1}{\sqrt3}\right)^2 \times 3 = \tfrac13 \times 3 = 1.

    So the length is 1. The vector is 133 PQ→\frac{1}{3\sqrt3}\,\overrightarrow{PQ}⁠, a positive multiple of PQ→\overrightarrow{PQ}⁠, so it points from P towards Q.

    X Y Z O P(1, 2, 3) Q(4, 5, 6) 3 3 3 PQ unit PQ = 3î + 3ĵ + 3k̂ |PQ| = 3√3 ≈ 5.20 unit vector: PQ/(3√3) = (î + ĵ + k̂)/√3 true lengths, one scale PQ, length 3√3 unit, length 1
    P(1, 2, 3) and Q(4, 5, 6) in a true (orthographic) view with one scale on all three axes. Going from P to Q means 3 units along x, 3 along y and 3 along z (dashed), so PQ→=3i^+3j^+3k^\overrightarrow{PQ} = 3\hat i + 3\hat j + 3\hat k⁠. The short arrow at P is the unit vector, 133\frac{1}{3\sqrt3} of PQ→\overrightarrow{PQ}⁠. In this view PQ→\overrightarrow{PQ} points nearly towards you and looks shorter than it is, so the inset on the right draws PQ→\overrightarrow{PQ} and the unit vector side by side at their true lengths, 333\sqrt3 and 1, with one scale.
    The direction matters: the question says "from P to Q" by writing PQ→\overrightarrow{PQ}⁠. The unit vector along QP→\overrightarrow{QP} would be the negative, −13(i^+j^+k^)-\frac{1}{\sqrt3}(\hat i + \hat j + \hat k)⁠.
    A unit vector has magnitude 1 (Section 10.3); λa⃗\lambda\vec a has the direction of a⃗\vec a for λ>0\lambda > 0 (Section 10.5). Compare Example 10, which finds such a vector PQ→\overrightarrow{PQ}⁠.
Result

The unit vector in the direction of PQ→\overrightarrow{PQ} is 13i^+13j^+13k^\frac{1}{\sqrt3}\hat i + \frac{1}{\sqrt3}\hat j + \frac{1}{\sqrt3}\hat k⁠.

(Textbook answer: 13i^+13j^+13k^\frac{1}{\sqrt3}\hat i + \frac{1}{\sqrt3}\hat j + \frac{1}{\sqrt3}\hat k⁠.)

Check it with numbers

∣PQ→∣=33≈5.196152|\overrightarrow{PQ}| = 3\sqrt3 \approx 5.196152⁠, the distance from P to Q. Each component of the unit vector is 13≈0.577350\frac{1}{\sqrt3} \approx 0.577350⁠.

The unit vector makes the same angle with every axis: cos⁡−1(0.577350)≈54.736∘\cos^{-1}(0.577350) \approx 54.736^\circ⁠.

Common mistakes
  1. Subtracting the wrong way. P minus Q gives −3i^−3j^−3k^-3\hat i - 3\hat j - 3\hat k⁠, the vector QP→\overrightarrow{QP}⁠, and its unit vector points the opposite way.
  2. Using a position vector instead. 177(4i^+5j^+6k^)\frac{1}{\sqrt{77}}(4\hat i + 5\hat j + 6\hat k) is the unit vector along OQ→\overrightarrow{OQ}⁠, not along PQ→\overrightarrow{PQ}⁠.
  3. Dividing by 27 or by 3. The length is 27=33\sqrt{27} = 3\sqrt3⁠; dividing by 3 gives i^+j^+k^\hat i + \hat j + \hat k⁠, whose length is 3\sqrt3⁠, not 1.
What to take away
  1. PQ→\overrightarrow{PQ} = (coordinates of Q) minus (coordinates of P).
  2. Unit vector along PQ→\overrightarrow{PQ} = PQ→\overrightarrow{PQ} divided by the distance PQ.
  3. Reversing the order of the points reverses the unit vector.

Every fact used, and where it comes from

FactUsed inSource
Vector joining two points; OP→+PQ→=OQ→\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}Step 1Section 10.5.2, Fig 10.15 (Basics: three numbers fix a point)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 2, 4Section 10.5.1
a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a⁠; λa⃗\lambda\vec a keeps the direction for λ>0\lambda > 0Steps 3, 4Section 10.5
λa⃗\lambda\vec a by componentsStep 3Section 10.5.1, (iv) (Lesson: components)
Unit vector: magnitude 1Step 4Section 10.3
The answerResultNCERT answer key, Exercise 10.2 Q8 (Lesson: components, lengths and unit vectors)
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →