Triangle Inequality (Bounding the Resultant of Two Vectors)
When you combine two displacements, two velocities, or two forces in "Motion in a Plane," you add them as vectors using the triangle law: place the tail of the second vector at the head of the first, and the resultant runs from the start to the finish. The triangle inequality is simply the statement that this resultant can never be longer than the two vectors laid end to end, and can never be shorter than their difference.
The Intuition
Suppose you walk 3 m in one direction, then 4 m in some other direction. Could you end up 8 m from where you started? No — the farthest you can possibly get is 3+4=7 m, and that only happens if both walks point the same way, so there's no bend at all (a "flat" triangle). The moment the second walk points in a different direction, a real corner appears, and cutting across that corner (the direct path) is always shorter than going via the corner. That's the geometric heart of every triangle: any one side is shorter than the sum of the other two, unless the triangle collapses onto a straight line.
The Precise Statement
For two vectors A and B added by the triangle law, the magnitude of the resultant R=A+B is bounded on both sides:
∣A∣−∣B∣≤∣A+B∣≤∣A∣+∣B∣
The upper bound∣A+B∣≤∣A∣+∣B∣ is reached only when A and B point in exactly the same direction (the angle between them is 0∘) — the triangle flattens out.
The lower bound∣A+B∣≥∣A∣−∣B∣ is reached only when A and B point in exactly opposite directions (the angle is 180∘).
For any angle in between, the resultant magnitude lies strictly between these two limits.
This is the vector form of the ordinary triangle inequality ∣x+y∣≤∣x∣+∣y∣ you may already know for numbers: here x and y become vectors, and "the sides of a triangle" become "a vector, another vector, and their sum."
Why It Matters in Kinematics
This bound is genuinely useful when combining physical quantities in the plane:
Relative velocity: if a boat has speed 5 m/s relative to water and the river flows at 3 m/s, the boat's speed relative to the ground must lie between ∣5−3∣=2 m/s and 5+3=8 m/s, depending on the angle the boat is steered — it can never be less than 2 or more than 8.
Combining forces or displacements: if you know only the magnitudes of two vectors, not the angle between them, this inequality instantly tells you the range of possible resultant magnitudes without doing any trigonometry. …
Concept: Vector triple products and the constraint A+B+C=0.
Since A+B+C=0, we have C=−(A+B). The three vectors form a closed triangle (or are collinear).
Check each option:
(A)(A×B)×C lies in the plane of A and B by the BAC-CAB rule. It is zero only when A×B=0 (i.e., A∥B) or when C is parallel to A×B (i.e., C is perpendicular to the plane of A, B). The condition "B, C parallel" is neither necessary nor sufficient. False.
(B)(A×B)⋅C is the scalar triple product, which equals zero if and only if the three vectors are coplanar. Since they sum to zero, they always lie in a plane (the triangle they form). Thus (A×B)⋅C=0 always, regardless of whether B∥C. False.
(C) As noted, (A×B)×C lies in the plane of A and B, which is the same plane containing all three vectors. True. …
Since A+B+C=0 the three vectors are coplanar, so the scalar triple product (A×B)⋅C is identically zero. Statement (B) claims it is non-zero (unless B∥C), which is wrong. (B) is the false statement.
The key consequence of A+B+C=0
Three vectors summing to zero form a closed triangle, so they lie in one plane - they are coplanar. Also C=−(A+B).
Statement (B) - the false one
(A×B)⋅C is the scalar triple product, the volume of the parallelepiped on A,B,C. For coplanar vectors this volume is zero. Directly:
(A×B)⋅C=(A×B)⋅[−(A+B)]=−(A×B)⋅A−(A×B)⋅B=0,
because A×B is perpendicular to both A and B. It is always zero, not "non-zero unless B,C are parallel." Hence (B) is false.
Why the others are true
(A) By the triple-product identity (A×B)×C=B(A⋅C)−A(B⋅C). Its magnitude equals ∣A×B∣∣C∣ (as A×B⊥C), which vanishes only when A∥B; but with A+B+C=0 that forces all three collinear, i.e. B∥C as well. So the statement holds - (A) is true. …
Concept: Settle the Triple Products with One Concrete Coplanar Triangle
Method: Explicit Coordinate Construction (Pick Real Numbers for A,B,C), Not Symbolic BAC-CAB Expansion
Rather than expanding (A×B)×C symbolically via the BAC-CAB identity, this method picks one concrete, easy-to-visualize triple of vectors satisfying A+B+C=0 and computes every quantity in all four options directly as numbers -- turning an abstract vector-identity question into ordinary arithmetic.
Step 1 -- Build an explicit example
Since any three vectors summing to zero form a closed triangle, take a simple right triangle in the xy-plane:
A=3i^,B=3j^,C=−3i^−3j^
Check: A+B+C=(3−3)i^+(3−3)j^=0. Good -- this is a legitimate instance of the problem's hypothesis, and the three vectors are visibly coplanar (all lie in the xy-plane, with k^-component zero).
Step 2 -- Compute (A×B) once, reuse it in every option
A×B=(3i^)×(3j^)=9(i^×j^)=9k^
Step 3 -- Test option (B): (A×B)⋅C
9k^⋅(−3i^−3j^)=9(0)=0
The claim in (B) is that this is non-zero unless B,C are parallel -- but here B=3j^ and C=−3i^−3j^ are clearly not parallel (different directions), and yet the scalar triple product is exactly 0. This single computed number directly contradicts (B). (B) is false -- confirmed by direct arithmetic, no identity needed.
This is manifestly non-zero, and B=3j^, C=−3i^−3j^ are not parallel in this example -- consistent with (A)'s claim.
Step 5 -- Test option (C): does (A×B)×C=27i^−27j^ lie in the plane of A,B,C?
All three original vectors lie in the xy-plane (zero k^-component); the computed result 27i^−27j^ also has zero k^-component. It lies in the same plane, consistent with (C)'s claim.