Vector Algebra JEE 2024: Minimum of 27|c − a|^2

What is the JEE Main 2024 vector algebra question with a = b × c and angle constraint?

Three vectors a, b, c satisfy |a| = 2, |b| = 3 and a = b × c. The angle α between b and c lies in the closed interval [0, π/3]. The minimum value of 27|c − a|^2 must be found.

How do you solve the official JEE Main 2024 vector algebra question step by step?

The minimum value is 124.

Given a = b × c, it follows that a · c = 0 from the scalar triple product property.

Thus

because the dot product vanishes.

Therefore

From |a| = 2 we have |b × c| = 2, so $3|\overrightarrow{c}|\sin \alpha =2$ hence

Then

The overall expression becomes

Inside [0, π/3], sin α is maximum at α = π/3 where sin²α = 3/4. The minimum of 12/sin²α equals 12/(3/4) = 16. Thus the minimum value is 108 + 16 = 124.

Why does skipping perpendicularity produce 121 instead of 124?

Skipping the perpendicularity step or mishandling the interval produces 121 instead of the correct minimum 124.

Students who erroneously retain a non-zero a · c term while expanding |c − a|^2 head toward the wrong number. They also treat the maximum of sin α as occurring outside the given interval or use sin(π/2) instead of sin(π/3).

Compare the expansions:

• Correct: 27|c|^2 + 108
• Incorrect: 27|c|^2 + 108 − 54 a·c term

The extra term arises only when the perpendicularity step is skipped.

Which vector properties must you recall instantly in these questions?

b × c is perpendicular to both b and c, hence a · c = 0 and a · b = 0. The magnitude relation is

The |u − v|^2 formula changes sharply depending on the dot product:

• When u · v = 0: |u − v|^2 = |u|^2 + |v|^2
• When the dot product is retained: |u − v|^2 = |u|^2 + |v|^2 − 2 u · v

These facts alone carry the solution. Write them first.

How does the angle interval [0, π/3] shift the minimum value?

The interval [0, π/3] shifts the minimum of 12/sin²α + 108 from 120 to 124.

Sin α is strictly increasing on [0, π/3]. The largest sin α inside the interval is √3/2 at α = π/3, giving sin²α = 3/4.

If the interval extended to π/2 then sin²α = 1 would give 12/1 + 108 = 120, which is lower but disallowed. The actual minimum therefore uses 16 in the reciprocal-sine term instead of 12.

The closed endpoint at π/3 forces the 16.

What two similar vector algebra questions test the same perpendicularity and interval skills?

Here are two practice problems that require explicit use of dot-product zero and sin-α maximisation inside an interval.

Question A: Given |a| = 3, a = b × c, |b| = 4, angle between b and c in [π/6, π/2], find the minimum of 16|c − a|^2.

Question B: Vectors p, q, r with p = q × r, |p| = 5, |q| = 2, find the minimum of |r − p/2|^2 when the angle between q and r ≤ π/4.

How can you finish hard vector algebra questions inside 300 seconds?

First line: write a · c = 0 and a · b = 0 from definition.

Second line: expand the square without the dot term.

Third line: substitute |c| from cross-product magnitude.

Final two lines: locate maximum of sin α inside the stated interval only.

This sequence keeps the calculation under two minutes. You can search every JEE Main paper from 2002 by chapter inside the past-paper archive to find more examples with worked solutions. If a mock test version still trips you up, photograph a doubt for a step-by-step reply.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

For a worked example of the same idea, see Trigonometric Functions JEE 2025: Inverse Equation Has Three Solutions.