How to find the magnitude of a cross product?

|a×b|² = |a|²|b|² - (a·b)². Alternatively, compute each component and sum squares.

What is the Cauchy-Schwarz inequality for vectors?

(c·d)² ≤ |c|²|d|². Equality holds when c and d are parallel (proportional vectors).

What is a vector in the yz-plane?

A vector d = (0, y, z) has its x-component zero, so it lies entirely in the yz-plane.

How to maximize a dot product?

c·d is maximized when d is in the direction of c. Maximum value = |c||d|. So max (c·d)² = |c|²|d|².

How to solve quadratic equations from magnitude condition?

Set |c|² equal to given value. Expand, collect terms, and solve using the quadratic formula or factoring.

Why is λ = -3 chosen over other roots?

The problem states λ is an integer. Solving 5λ²+4λ-33=0 gives λ=3 or λ=-3. Both may be checked; the problem typically specifies the negative integer root.

What does (c·d)² represent geometrically?

It is the square of the projection of d onto c, scaled by |c|. Maximized when c and d are parallel.

How to verify the answer 208?

With λ=-3, c=(1,-4,-6), |c|²=1+16+36=53. With d in yz-plane |d|=2: max (c·d)² = |c|²|d|² = 53×4 = 212? No: c·d = -4y-6z, max² = (16+36)(y²+z²)=52×4=208.

What is the difference between |c|² and the Cauchy-Schwarz bound?

|c|² = 53 uses the full c vector. But c·d only involves the yz-components of c (since d has no x-component), giving bound ((-4)²+(-6)²)|d|² = 52×4 = 208.

Let a = 2i - j + k, b = λj + 2k. If c = a × b and |c| = √53, then (c

Q: How do you compute a cross product?

For a = (a1,a2,a3) and b = (b1,b2,b3), c = a×b uses the determinant formula with i,j,k unit vectors. c = (a2b3-a3b2)i - (a1b3-a3b1)j + (a1b2-a2b1)k.

Let a = 2i – j + k, b = λj + 2k. If c = a × b and |c| = √53, then (c · d)² equals | JEE Main Mathematics

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Math Vectors

Q MCQ Vectors

Let $\mathbf{a} = 2\hat{i} – \hat{j} + \hat{k}$, $\mathbf{b} = \lambda\hat{j} + 2\hat{k}$. If $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ and $|\mathbf{c}| = \sqrt{53}$, where $\lambda$ is an integer, and $\mathbf{d}$ is a vector in the $yz$-plane with $|\mathbf{d}| = 2$, then the maximum value of $(\mathbf{c} \cdot \mathbf{d})^2$ is

A) 148

B) 196

C) 172

D) 208 ✓

✅ Correct Answer

D) 208

✓

Solution Steps

Write the vectors

$\mathbf{a} = (2,\,-1,\,1)$, $\mathbf{b} = (0,\,\lambda,\,2)$

We need $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ with $|\mathbf{c}| = \sqrt{53}$

Compute the cross product

$$\mathbf{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-1&1\\0&\lambda&2\end{vmatrix}$$

$c_x = (-1)(2) – (1)(\lambda) = -2 – \lambda$

$c_y = -[(2)(2) – (1)(0)] = -4$

$c_z = (2)(\lambda) – (-1)(0) = 2\lambda$

So $\mathbf{c} = (-2-\lambda,\; -4,\; 2\lambda)$

Apply the magnitude condition

$|\mathbf{c}|^2 = (\lambda+2)^2 + 16 + 4\lambda^2 = 53$

$\lambda^2 + 4\lambda + 4 + 16 + 4\lambda^2 = 53$

$5\lambda^2 + 4\lambda + 20 = 53$

$5\lambda^2 + 4\lambda – 33 = 0$

Solve for integer λ

$\lambda = \dfrac{-4 \pm \sqrt{16 + 660}}{10} = \dfrac{-4 \pm \sqrt{676}}{10} = \dfrac{-4 \pm 26}{10}$

$\lambda = \dfrac{22}{10} = 2.2$ (not integer) or $\lambda = \dfrac{-30}{10} = -3$ ✓

Since $\lambda$ must be an integer: $\lambda = -3$

Find vector c with λ = −3

$\mathbf{c} = (-2-(-3),\;-4,\;2(-3)) = (1,\;-4,\;-6)$

Verify: $|c|^2 = 1 + 16 + 36 = 53$ ✓

Set up d in the yz-plane

$\mathbf{d} = (0,\;y,\;z)$ with $y^2 + z^2 = 4$ (since $|\mathbf{d}| = 2$)

$\mathbf{c} \cdot \mathbf{d} = (1)(0) + (-4)(y) + (-6)(z) = -4y – 6z$

Maximize using Cauchy-Schwarz

By Cauchy-Schwarz inequality:

$(-4y – 6z)^2 \leq ((-4)^2 + (-6)^2)(y^2 + z^2)$

$(\mathbf{c} \cdot \mathbf{d})^2 \leq (16 + 36)(4) = 52 \times 4 = \mathbf{208}$

Equality when $(y,z) \propto (4,6)$, which satisfies $y^2+z^2=4$.

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Key Insight

Since d has no x-component, only the yz-components of c matter. The effective |c|² for the dot product bound is 4² + 6² = 52, not 53. So max (c·d)² = 52 × 4 = 208.

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Vectors Theory

1. Cross Product Formula

For vectors $\mathbf{a} = (a_1,a_2,a_3)$ and $\mathbf{b} = (b_1,b_2,b_3)$, the cross product $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ is computed using the determinant: $c_x = a_2b_3 – a_3b_2$, $c_y = a_3b_1 – a_1b_3$, $c_z = a_1b_2 – a_2b_1$. The result is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. The magnitude $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$ where $\theta$ is the angle between them.

2. Dot Product and Projection

The dot product $\mathbf{c} \cdot \mathbf{d} = c_1d_1 + c_2d_2 + c_3d_3 = |\mathbf{c}||\mathbf{d}|\cos\theta$. When $\mathbf{d}$ is restricted to the yz-plane ($d_x = 0$), only the $y$ and $z$ components of $\mathbf{c}$ contribute. The maximum dot product is $|\mathbf{c}||\mathbf{d}|$ when vectors are parallel, but here d is constrained to yz-plane so only the yz-projection of c is used.

3. Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality states $(a_1b_1 + a_2b_2)^2 \leq (a_1^2 + a_2^2)(b_1^2 + b_2^2)$. For our problem: $(-4y-6z)^2 \leq (16+36)(y^2+z^2)$. Equality holds when $\frac{y}{-4} = \frac{z}{-6}$, i.e., $y:z = 4:6 = 2:3$. This gives the configuration that maximizes the dot product.

4. Integer Constraints in Quadratics

When solving $5\lambda^2 + 4\lambda – 33 = 0$, the discriminant is $\Delta = 16 + 660 = 676 = 26^2$. Roots: $\lambda = \frac{-4 \pm 26}{10}$. One root is $\lambda = 2.2$ (not integer) and other is $\lambda = -3$ (integer). Always verify by substituting back: $5(9) + 4(-3) – 33 = 45 – 12 – 33 = 0$ ✓.

5. Vectors in Coordinate Planes

A vector in the yz-plane has the form $\mathbf{d} = (0, y, z)$. Its magnitude is $\sqrt{y^2+z^2}$. When computing $\mathbf{c} \cdot \mathbf{d}$, the x-component of $\mathbf{c}$ makes no contribution. This effectively reduces a 3D problem to 2D, simplifying the optimization. The constraint $|\mathbf{d}|=2$ means we are optimizing on a circle of radius 2 in the yz-plane.

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FAQs

How do you compute a cross product?

⌄

Use the determinant: c = a×b gives cx = a2b3-a3b2, cy = a3b1-a1b3, cz = a1b2-a2b1. Result is perpendicular to both a and b.

How to find magnitude of cross product?

⌄

|a×b|² = sum of squares of all three components. Here: |c|² = (λ+2)² + 16 + 4λ². Set equal to 53 and solve for λ.

What is the Cauchy-Schwarz inequality?

⌄

(a·b)² ≤ |a|²|b|². For yz-plane: (-4y-6z)² ≤ (16+36)(y²+z²) = 52×4 = 208. Equality when y/4 = z/6.

Why does d in yz-plane matter?

⌄

d = (0,y,z) so c·d = -4y-6z (x-component cancels). The bound uses only yz-components of c, giving 4²+6²=52, not |c|²=53.

Why is λ = -3 chosen?

⌄

Solving 5λ²+4λ-33=0 gives λ=2.2 (not integer) or λ=-3 (integer). Since the problem specifies integer λ, we take λ=-3.

How to verify |c|²=53 with λ=-3?

⌄

c = (1,-4,-6): |c|² = 1+16+36 = 53 ✓. All three components confirmed correct.

What d vector achieves the maximum?

⌄

d proportional to (0,4,6) normalized to |d|=2. So d = (0, 4/√13 × 2/√4, …) — the direction (0,2,3)/√13 × 2.

Why is the answer 208 not 212?

⌄

If d were unrestricted in 3D: max (c·d)² = |c|²|d|² = 53×4=212. But d is in yz-plane, so only yz-part of c (4²+6²=52) counts: 52×4=208.

What does |d|=2 constraint mean?

⌄

y²+z²=4 confines d to a circle of radius 2 in the yz-plane. We maximize (c·d)² = (-4y-6z)² over this circle.

When does equality hold in Cauchy-Schwarz?

⌄

When vectors are proportional: (y,z) ∝ (4,6). Normalize: y=4/√13, z=6/√13, but scale so y²+z²=4, giving y=8/√13, z=12/√13.

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