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πŸ”¬ Science❓ Question

Cross product: I can calculate it but I never know which way the answer points

Ananya Reddy11th Sample

14 Oct 2026 Β· 2 min read

I'm in class 11 (PCB) and we just started torque and angular momentum. I can do the arithmetic for the cross product, but I keep losing marks on direction and I think my understanding is a bit broken.

What I know: Ο„βƒ—=rβƒ—Γ—Fβƒ—\vec\tau = \vec r \times \vec F, and the magnitude is rFsin⁑θrF\sin\theta. Direction is "perpendicular to both" by the right-hand rule.

Here is where I get confused:

  1. Example from my notes: r⃗=2i^\vec r = 2\hat i m and F⃗=3j^\vec F = 3\hat j N. I got τ⃗=6k^\vec \tau = 6\hat k N m. Good. But my notes also say this means the wheel rotates "anticlockwise", and I can't see how an arrow pointing along k^\hat k is related to anticlockwise. Where is k^\hat k even pointing on my page, into the page or out of it?
  2. I once wrote a⃗×b⃗=b⃗×a⃗\vec a \times \vec b = \vec b \times \vec a in a test because dot product is like that. I'm now unsure whether the sign flips or not.
  3. In the board syllabus we also use Fleming's left-hand rule for the force on a current-carrying wire. Is that a different rule from the right-hand rule, or the same thing? Do I need to remember both?

For a practice problem I tried aβƒ—=i^+2j^+3k^\vec a = \hat i + 2\hat j + 3\hat k and bβƒ—=4i^+5j^+6k^\vec b = 4\hat i + 5\hat j + 6\hat k using a determinant and got (3,βˆ’6,3)(3, -6, 3). The book says (βˆ’3,6,βˆ’3)(-3, 6, -3). I've checked three times and I can't find my mistake.

Please explain it like I'm not good at maths, I'm much better at biology than at visualising 3D.

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9 answers & replies

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  • [+]Neel Chatterjee12thSampleΒ· 6d ago

    Start with this and the rest follows. Put your page flat on the desk, i^\hat i pointing right, j^\hat j pointing up the page (away from you), then k^\hat k points out of the page toward your face. That's the standard orientation.

    Now take your fingers, point them along i^\hat i and curl them toward j^\hat j (the shorter way, 90 degrees). Your thumb points toward your face, which is +k^+\hat k. So i^Γ—j^=+k^\hat i \times \hat j = +\hat k, and rotating from i^\hat i toward j^\hat j is what? Right to up. That's anticlockwise when you look at the page. So "torque along +k^+\hat k" and "anticlockwise" are the same statement. The vector direction is the axis of rotation, and the right-hand rule tells you which sense along that axis goes with which spin.

  • [+]Rohan Sharma11thSampleΒ· 6d ago

    for point 2, aΓ—b = bΓ—a? no wait. cross product is cosΞΈ so if you swap them nothing changes

    • [+]Tushar PawarDropperSampleΒ· 6d ago

      Careful, two things wrong there. Cross product magnitude uses sinΞΈ (that's what vanishes when vectors are parallel), cosΞΈ is the dot product. And the order matters: aβƒ—Γ—bβƒ—=βˆ’bβƒ—Γ—aβƒ—\vec a\times\vec b = -\vec b\times\vec a. Swap the order and the direction flips. Magnitude stays the same. Dot product is the one that doesn't care about order.

      • [+]Rohan Sharma11thSampleΒ· 6d ago

        oh. yeah I mixed those 😭 sin for cross, cos for dot. "cross = sin" like the letters are both... ok no mnemonic, just memorising

  • [+]Nisha Verma12thSampleΒ· 6d ago

    About your numbers: your determinant is probably right but you wrote the sign wrong somewhere. Do it by components instead, it's less error-prone than a 3x3 determinant when you're tired:

    (aybzβˆ’azby,Β azbxβˆ’axbz,Β axbyβˆ’aybx)(a_yb_z - a_zb_y,\ a_zb_x - a_xb_z,\ a_xb_y - a_yb_x)

    For a⃗=(1,2,3)\vec a = (1,2,3), b⃗=(4,5,6)\vec b = (4,5,6):

    • x: 2β‹…6βˆ’3β‹…5=12βˆ’15=βˆ’32\cdot 6 - 3\cdot 5 = 12 - 15 = -3
    • y: 3β‹…4βˆ’1β‹…6=12βˆ’6=63\cdot 4 - 1\cdot 6 = 12 - 6 = 6
    • z: 1β‹…5βˆ’2β‹…4=5βˆ’8=βˆ’31\cdot 5 - 2\cdot 4 = 5 - 8 = -3

    So (βˆ’3,6,βˆ’3)(-3, 6, -3), same as the book. Your (3,βˆ’6,3)(3,-6,3) is exactly bβƒ—Γ—aβƒ—\vec b\times\vec a. I bet the middle term of your determinant has a sign error. People forget the minus in front of the j^\hat j term of the expansion.

    • [+]Ananya Reddy11thSampleΒ· 6d agoAuthor

      yes!! I did the j-term with a plus. Thank you, I redid it and got -3, 6, -3. I'm going to colour code the signs in my notes πŸ˜…

  • [+]Siddharth RaoDropperSampleΒ· 5d ago

    On Fleming. It's not a different rule, it's the same physics with a different hand trick. The force on a current-carrying wire is Fβƒ—=I Lβƒ—Γ—Bβƒ—\vec F = I\,\vec L\times\vec B. Right-hand rule for cross products: fingers along Lβƒ—\vec L, curl toward Bβƒ—\vec B, thumb gives Fβƒ—\vec F. Fleming's left hand: thumb = force, first finger = field, second finger = current, all mutually perpendicular. You get the same direction both ways. I just stuck to the cross-product way in my drop year so I only had one rule to remember. Use whichever one your teacher accepts in the boards, but know that they agree.

    • [+]Tanvi Deshmukh12thSampleΒ· 5d ago

      this!! I made a little table in my notes: iΓ—j = k, jΓ—k = i, kΓ—i = j (cyclic order), and reverse order gives the minus sign. then Fleming is just for the exam if they ask "state the rule" πŸ™‚

      • [+]Ananya Reddy11thSampleΒ· 5d agoAuthor

        this finally makes sense. so k is out of the page, anticlockwise = +k, and swapping order flips the sign. I'll recheck my torque examples tonight.

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