Three Dimensional Geometry JEE 2009: Matrix-Match Solution

Q: How do I calculate the parallelepiped volume in row C?

For the common-vertex edges (1,1,0), (1,2,0) and (1,1,π), take the absolute value of their scalar triple product. Expanding the determinant down the third column gives V = |π(2 − 1)| = π. Therefore, row C matches t.

Q: How do I find the angle between the unit vectors in row D?

From a + b + √3c = 0 with all three vectors of unit magnitude, obtain |a + b|² = 3. Expanding gives 1 + 1 + 2a·b = 3, so cos θ = a·b = 1/2. The angle is π/3, and row D matches r.

What is the matching answer for Three Dimensional Geometry JEE 2009?

The Three Dimensional Geometry JEE 2009 matching key is A–q,s; B–p,r,s,t; C–t; D–r. The volume row describes a parallelepiped formed by three edge vectors drawn from a common vertex, so its volume comes from their scalar triple product.

A parallelepiped on coordinate axes x, y and z with common vertex O(0,0,0), adjacent vertices U(1,1,0), V(1,2,0) and W(1,1,π), label its directed edges OU=u, OV=v and OW=w, and complete the remaining edges by parallel translation.

This is a JEE 2009 Paper 2, Mathematics, matrix-match question, classified under Three Dimensional Geometry. Its four rows also test trigonometry and continuity.

Match each statement with every suitable candidate value:

A: Select the listed angles satisfying

2 \sin^{2} θ + \sin^{2} (2 θ) = 2 .

B: Select the listed discontinuities of

f (x) = ⌊ \frac{6 x}{π} ⌋ \cos (⌊ \frac{3 x}{π} ⌋) .

Floor means the greatest integer not exceeding its argument. Cosine uses radians and acts on the second floor value.

C: Find the parallelepiped’s volume for the edge vectors

\hat{i} + \hat{j}, \hat{i} + 2 \hat{j}, \hat{i} + \hat{j} + π \hat{k} .

D: Find the angle between the first two unit vectors when

\vec{a} + \vec{b} + \sqrt{3} \vec{c} = \vec{0}, | \vec{a} | = | \vec{b} | = | \vec{c} | = 1 .

The candidate key is:

p: $\frac{π}{6}$
q: $\frac{π}{4}$
r: $\frac{π}{3}$
s: $\frac{π}{2}$
t: $π$

A row can match multiple values, and a value can serve multiple rows. Check each pairing independently.

Which listed angles satisfy row A?

Only q and s work among the supplied values. The official substitution method reduces the equation to a quadratic in sine squared.

Use the double-angle identity and substitution:

\sin^{2} (2 θ) = 4 \sin^{2} θ \cos^{2} θ, u = \sin^{2} θ, \cos^{2} θ = 1 - u .

Then:

2 u + 4 u (1 - u) & = 2 6 u - 4 u^{2} & = 2 4 u^{2} - 6 u + 2 & = 0 2 u^{2} - 3 u + 1 & = 0 (2 u - 1) (u - 1) & = 0 .

u = \frac{1}{2} or u = 1 .

Compare all five candidates:

p, reject:

\sin^{2} \frac{π}{6} = \frac{1}{4} .

q, include:

\sin^{2} \frac{π}{4} = \frac{1}{2} .

r, reject:

\sin^{2} \frac{π}{3} = \frac{3}{4} .

s, include:

\sin^{2} \frac{π}{2} = 1 .

t, reject: $\sin^{2} π = 0 .$

Therefore, A–q,s. These are the matches among the supplied values, not the equation’s only roots over all real angles.

Where is the floor-function product discontinuous in row B?

The product is discontinuous at p, r, s and t, but continuous at q. Floor boundaries identify possible jumps; the product’s actual one-sided limits must still be compared.

The possible jump locations satisfy:

\frac{6 x}{π} \in Z ⟹ x = \frac{k π}{6}, \frac{3 x}{π} \in Z ⟹ x = \frac{k π}{3}, k \in Z .

Each ordered pair below records the first and second floor values. The arrows run from immediately left to immediately right.

p, include: Floor pairs change from

x = \frac{π}{6} : (0, 0) ⟶ (1, 0) .

The left-hand limit and right-hand limit/value are

f (x^{-}) = 0, f (x^{+}) = f (x) = 1 .

q, exclude: Both floors stay fixed in a neighbourhood:

x = \frac{π}{4} : (1, 0) ⟶ (1, 0),

f (x^{-}) = 1, f (x^{+}) = f (x) = 1 .

r, include: The limits are close, but unequal:

x = \frac{π}{3} : (1, 0) ⟶ (2, 1),

f (x^{-}) = 1, f (x^{+}) = f (x) = 2 \cos (1) \approx 1.0806 .

s, include: Only the first floor changes:

x = \frac{π}{2} : (2, 1) ⟶ (3, 1),

f (x^{-}) = 2 \cos (1), f (x^{+}) = f (x) = 3 \cos (1) .

The limits differ because $\cos (1) \neq 0 .$

t, include: Both floors change:

x = π : (5, 2) ⟶ (6, 3),

f (x^{-}) = 5 \cos (2) \approx - 2.0807,

f (x^{+}) = f (x) = 6 \cos (3) \approx - 5.9400 .

At every candidate, the function value equals the right-hand limit: each floor is right-continuous, and so is their stated product. The unequal one-sided limits establish B–p,r,s,t.

How do I calculate the parallelepiped’s volume in row C?

The volume matches t. Use the three pictured edges from the common vertex as the determinant rows, following the official scalar-triple-product method.

\vec{u} = (1, 1, 0), \vec{v} = (1, 2, 0), \vec{w} = (1, 1, π) .

V = | \vec{u} \cdot (\vec{v} \times \vec{w}) | = | | \begin{matrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & π \end{matrix} | | .

Expand down the third column:

V = | π | \begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix} | | = | π (1 \times 2 - 1 \times 1) | = | π | = π .

The absolute value converts the signed scalar triple product into non-negative volume, giving C–t. For optional determinant practice, use Matrices and Determinants JEE 2010: Worked Solution.

How do I find the angle between the unit vectors in row D?

The angle matches r. Rearrange the relation, take magnitudes and square to expose the dot product without finding vector components. $\vec{a} + \vec{b} = - \sqrt{3} \vec{c},$

| \vec{a} + \vec{b} | = \sqrt{3} | \vec{c} | = \sqrt{3} .

Squaring and expanding gives: $| \vec{a} + \vec{b} |^{2} = 3,$

| \vec{a} |^{2} + | \vec{b} |^{2} + 2 \vec{a} \cdot \vec{b} = 3 .

Using the unit magnitudes:

1 + 1 + 2 \vec{a} \cdot \vec{b} = 3 ⟹ 2 \vec{a} \cdot \vec{b} = 1 ⟹ \vec{a} \cdot \vec{b} = \frac{1}{2} .

For the angle between vectors: $0 \leq θ \leq π,$

\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ = \cos θ = \frac{1}{2} ⟹ θ = \frac{π}{3} .

Thus D–r. The complete matching key is:

A–q,s
B–p,r,s,t
C–t
D–r

Why does rounding wrongly add q to row B?

Nearest-integer rounding changes at half-integers; floor does not. Confusing these operations falsely suggests the incorrect row-value pairing B–q.

At q:

x = \frac{π}{4} ⟹ \frac{6 x}{π} = \frac{3}{2} .

Nearest-integer rounding changes across this half-integer. The actual floor stays fixed, as does the second floor, throughout this interval:

\frac{π}{6} \leq x < \frac{π}{3} : ⌊ \frac{6 x}{π} ⌋ = 1, ⌊ \frac{3 x}{π} ⌋ = 0 .

$f (x) = 1 \cos (0) = 1 .$ This constant interval contains q in its interior, proving continuity there. Repair the method: locate integer arguments, not half-integers, then check whether the full expression actually changes.

Which two related questions should I solve next?

Practise changing the third edge and the vector coefficient. Both exercises below are original related practice, not additional JEE past-paper questions.

Question 1: For these edges, find the volume in terms of the real parameter and the condition for coplanarity:

\vec{u} = (1, 1, 0), \vec{v} = (1, 2, 0), \vec{w} = (1, 1, λ) .

The same determinant expansion gives:

\vec{u} \cdot (\vec{v} \times \vec{w}) = λ (2 - 1) = λ .

V = | λ |, coplanar exactly when λ = 0 .

Question 2: Three unit vectors satisfy the relation below. Find the angle between the first two.

\vec{a} + \vec{b} + \sqrt{2} \vec{c} = \vec{0} .

Taking magnitudes and squaring gives:

| \vec{a} + \vec{b} |^{2} = 2 ⟹ 1 + 1 + 2 \cos θ = 2 ⟹ \cos θ = 0 ⟹ θ = \frac{π}{2} .

Cover both answers and reproduce the determinant expansion and magnitude-squaring steps before checking your result.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the matching answer for Three Dimensional Geometry JEE 2009?

The matching key is A–q,s; B–p,r,s,t; C–t; D–r. The candidate values are p = π/6, q = π/4, r = π/3, s = π/2 and t = π. Each row can match more than one value.

Why does q not match row B in the JEE 2009 question?

At q = π/4, the function f(x) = ⌊6x/π⌋ cos(⌊3x/π⌋) is continuous because both floor values remain fixed nearby. Throughout π/6 ≤ x < π/3, they equal 1 and 0 respectively, so f(x) = 1. Nearest-integer rounding would incorrectly introduce a jump at π/4; the floor function does not.

How do I calculate the parallelepiped volume in row C?

For the common-vertex edges (1,1,0), (1,2,0) and (1,1,π), take the absolute value of their scalar triple product. Expanding the determinant down the third column gives V = |π(2 − 1)| = π. Therefore, row C matches t.

How do I find the angle between the unit vectors in row D?

From a + b + √3c = 0 with all three vectors of unit magnitude, obtain |a + b|² = 3. Expanding gives 1 + 1 + 2a·b = 3, so cos θ = a·b = 1/2. The angle is π/3, and row D matches r.

Three Dimensional Geometry JEE 2009: Matrix-Match Solution

What is the matching answer for Three Dimensional Geometry JEE 2009?

Which listed angles satisfy row A?

Where is the floor-function product discontinuous in row B?

How do I calculate the parallelepiped’s volume in row C?

How do I find the angle between the unit vectors in row D?

Why does rounding wrongly add q to row B?

Which two related questions should I solve next?

Frequently asked questions

What is the matching answer for Three Dimensional Geometry JEE 2009?

Why does q not match row B in the JEE 2009 question?

How do I calculate the parallelepiped volume in row C?

How do I find the angle between the unit vectors in row D?

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