Logo
Jeeify
NVAMediumJEE 2025Cross Product

JEE Mathematics 2025 Question with Solution

Let a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}, b⃗=3i^+2j^−k^\vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}, c⃗=λj^+μk^\vec{c} = \lambda \hat{j} + \mu \hat{k} and d^\hat{d} be a unit vector such that a⃗×d^=b⃗×d^\vec{a} \times \hat{d} = \vec{b} \times \hat{d} and c⃗⋅d^=1\vec{c} \cdot \hat{d} = 1. If c⃗\vec{c} is perpendicular to a⃗\vec{a}, then ∣3λd^+μc⃗∣2|3\lambda \hat{d} + \mu \vec{c}|^2 is equal to _____.

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}, b⃗=3i^+2j^−k^\vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}, c⃗=λj^+μk^\vec{c} = \lambda \hat{j} + \mu \hat{k}, and d^\hat{d} is a unit vector such that a⃗×d^=b⃗×d^\vec{a} \times \hat{d} = \vec{b} \times \hat{d} and c⃗⋅d^=1\vec{c} \cdot \hat{d} = 1.

Find: ∣3λd^+μc⃗∣2|3\lambda \hat{d} + \mu \vec{c}|^2.

From

a⃗×d^=b⃗×d^\vec{a} \times \hat{d} = \vec{b} \times \hat{d}

we get

a⃗×d^−b⃗×d^=0\vec{a} \times \hat{d} - \vec{b} \times \hat{d} = 0 (a⃗−b⃗)×d^=0(\vec{a} - \vec{b}) \times \hat{d} = 0

So, a⃗−b⃗\vec{a} - \vec{b} is parallel to d^\hat{d}.

Now,

a⃗−b⃗=(i^+j^+k^)−(3i^+2j^−k^)=−2i^−j^+2k^\vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (3\hat{i} + 2\hat{j} - \hat{k}) = -2\hat{i} - \hat{j} + 2\hat{k}

Hence,

d^=a⃗−b⃗∣a⃗−b⃗∣=−2i^−j^+2k^(−2)2+(−1)2+22=−2i^−j^+2k^3\hat{d} = \frac{\vec{a} - \vec{b}}{|\vec{a} - \vec{b}|} = \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{(-2)^2 + (-1)^2 + 2^2}} = \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}

Using c⃗⋅d^=1\vec{c} \cdot \hat{d} = 1,

(λj^+μk^)⋅−2i^−j^+2k^3=1(\lambda \hat{j} + \mu \hat{k}) \cdot \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} = 1 −λ+2μ3=1\frac{-\lambda + 2\mu}{3} = 1 −λ+2μ=3-\lambda + 2\mu = 3

Also, since c⃗\vec{c} is perpendicular to a⃗\vec{a},

c⃗⋅a⃗=0\vec{c} \cdot \vec{a} = 0 (λj^+μk^)⋅(i^+j^+k^)=0(\lambda \hat{j} + \mu \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0 λ+μ=0⇒μ=−λ\lambda + \mu = 0 \Rightarrow \mu = -\lambda

Substituting μ=−λ\mu = -\lambda into −λ+2μ=3-\lambda + 2\mu = 3,

−λ−2λ=3-\lambda - 2\lambda = 3 −3λ=3⇒λ=−1,μ=1-3\lambda = 3 \Rightarrow \lambda = -1, \quad \mu = 1

Therefore,

c⃗=−j^+k^\vec{c} = -\hat{j} + \hat{k}

Now,

3λd^+μc⃗=3(−1)⋅−2i^−j^+2k^3+(−j^+k^)3\lambda \hat{d} + \mu \vec{c} = 3(-1) \cdot \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} + (-\hat{j} + \hat{k}) =2i^+j^−2k^−j^+k^=2i^−k^= 2\hat{i} + \hat{j} - 2\hat{k} - \hat{j} + \hat{k} = 2\hat{i} - \hat{k}

Hence,

∣3λd^+μc⃗∣2=∣2i^−k^∣2=22+(−1)2=4+1=5|3\lambda \hat{d} + \mu \vec{c}|^2 = |2\hat{i} - \hat{k}|^2 = 2^2 + (-1)^2 = 4 + 1 = 5

Therefore, the required value is 55.

Using component equations for the unit vector

Given: The same vector data and conditions.

Find: ∣3λd^+μc⃗∣2|3\lambda \hat{d} + \mu \vec{c}|^2.

Let

d^=d1i^+d2j^+d3k^\hat{d} = d_1\hat{i} + d_2\hat{j} + d_3\hat{k}

From a⃗×d^=b⃗×d^\vec{a} \times \hat{d} = \vec{b} \times \hat{d},

∣i^j^k^111d1d2d3∣=∣i^j^k^32−1d1d2d3∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ d_1 & d_2 & d_3 \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ d_1 & d_2 & d_3 \end{vmatrix}

This relation is equivalent to

(a⃗−b⃗)×d^=0(\vec{a}-\vec{b}) \times \hat{d} = 0

so d^\hat{d} must be parallel to −2i^−j^+2k^-2\hat{i} - \hat{j} + 2\hat{k}. Since d^\hat{d} is a unit vector,

d^=−2i^−j^+2k^3\hat{d} = \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}

Now use c⃗⋅d^=1\vec{c} \cdot \hat{d} = 1:

(λj^+μk^)⋅−2i^−j^+2k^3=1(\lambda \hat{j} + \mu \hat{k}) \cdot \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} = 1 −λ+2μ=3-\lambda + 2\mu = 3

And since c⃗⊥a⃗\vec{c} \perp \vec{a},

(λj^+μk^)⋅(i^+j^+k^)=0(\lambda \hat{j} + \mu \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0 λ+μ=0\lambda + \mu = 0

Solving these two equations gives

λ=−1,μ=1\lambda = -1, \quad \mu = 1

Thus,

3λd^+μc⃗=−3d^+c⃗=2i^−k^3\lambda \hat{d} + \mu \vec{c} = -3\hat{d} + \vec{c} = 2\hat{i} - \hat{k}

Therefore,

∣3λd^+μc⃗∣2=5|3\lambda \hat{d} + \mu \vec{c}|^2 = 5

So the answer is 55.

Common mistakes

  • Assuming directly that a⃗=b⃗\vec{a} = \vec{b} from a⃗×d^=b⃗×d^\vec{a} \times \hat{d} = \vec{b} \times \hat{d} is incorrect. The correct step is to write (a⃗−b⃗)×d^=0(\vec{a}-\vec{b}) \times \hat{d} = 0, which means a⃗−b⃗\vec{a}-\vec{b} is parallel to d^\hat{d}.

  • Forgetting that d^\hat{d} is a unit vector leads to using −2i^−j^+2k^-2\hat{i}-\hat{j}+2\hat{k} directly instead of normalizing it. You must divide by its magnitude 33 to get the correct d^\hat{d}.

  • Using the perpendicular condition incorrectly is a common error. Since c⃗⊥a⃗\vec{c} \perp \vec{a}, one must use the dot product equation c⃗⋅a⃗=0\vec{c} \cdot \vec{a} = 0, which gives λ+μ=0\lambda + \mu = 0, not any cross-product relation.

Practice more Cross Product questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Cross Product

Related questions