If x, y and z Are Three Unit Vectors in Three-Dimensional Space, Then the Minimum Value of |x+y|² + |y+z|² + |z+x|² Is

JEE Main 2014. Previous Year Question.

Problem

If $x, y$ and $z$ are three unit vectors in three-dimensional space, then the minimum value of $|x+y|^2+|y+z|^2+|z+x|^2$ is?

Key insight. Expanding each squared term brings in the pairwise dot products $x\cdot y$, $y\cdot z$, $z\cdot x$ — and their sum is exactly what shows up (scaled) inside $|x+y+z|^2$. Since a squared magnitude can never be negative, that single fact pins down how small the pairwise-dot-product sum can possibly be.

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Approach

Expand each of the three squared terms using $|u+v|^2=|u|^2+|v|^2+2u\cdot v$, and add them up — this expresses the sum in terms of $x\cdot y+y\cdot z+z\cdot x$. Separately, expand $|x+y+z|^2\geq0$ to get a lower bound on that same pairwise-dot-product sum. Combine the two to get the minimum of the original expression.

Solution

Step 1 — Expand each squared term

$$|x+y|^2 = |x|^2+|y|^2+2\,x\cdot y = 2+2\,x\cdot y$$

(since $x,y,z$ are unit vectors, $|x|^2=|y|^2=|z|^2=1$)

Similarly:

$$|y+z|^2 = 2+2\,y\cdot z, \qquad |z+x|^2 = 2+2\,z\cdot x$$

Step 2 — Add all three

$$|x+y|^2+|y+z|^2+|z+x|^2 = 6+2(x\cdot y+y\cdot z+z\cdot x)$$

Step 3 — Find a lower bound on x·y + y·z + z·x using |x+y+z|²≥0

$$|x+y+z|^2 = |x|^2+|y|^2+|z|^2+2(x\cdot y+y\cdot z+z\cdot x) = 3+2(x\cdot y+y\cdot z+z\cdot x)$$

Since a squared magnitude can never be negative:

$$3+2(x\cdot y+y\cdot z+z\cdot x) \geq 0 \implies x\cdot y+y\cdot z+z\cdot x \geq -\frac32$$

Step 4 — Substitute this bound into Step 2

$$|x+y|^2+|y+z|^2+|z+x|^2 \geq 6+2\left(-\frac32\right) = 6-3 = 3$$

Step 5 — Confirm the bound is achievable

Equality requires $|x+y+z|^2=0$, i.e. $x+y+z=\vec0$ — achievable, for instance, by three unit vectors in a plane at $120°$ to each other (each pair then has $\cos120°=-\tfrac12$, giving $x\cdot y+y\cdot z+z\cdot x=-\tfrac32$ exactly). So the minimum is genuinely attained, not just a theoretical bound.

Answer

Common mistakes

• Treating $x\cdot y$, $y\cdot z$, $z\cdot x$ as independent quantities that can each be minimised separately. They’re linked through the vectors’ geometry — pushing one pairwise dot product very negative constrains what the others can be, which is exactly why the $|x+y+z|^2\geq0$ trick is needed rather than naive term-by-term minimisation.
• Forgetting to verify the bound is achievable. An inequality alone only gives a lower bound — confirming that some actual configuration of unit vectors (like three at $120°$ in a plane) attains equality is what turns “at least $3$” into “the minimum is exactly $3$.”

Practise next

• If $p,q,r$ are three unit vectors in 3D space, find the maximum value of $|p-q|^2+|q-r|^2+|r-p|^2$, using the same expansion approach (this time via $|p+q+r|^2\leq$ its own natural upper bound).