Logo
Jeeify
NVAMediumJEE 2025Cross Product

JEE Mathematics 2025 Question with Solution

Let a⃗=i^+2j^+k^\vec{a} = \hat{i} + 2\hat{j} + \hat{k}, b⃗=3i^−3j^+3k^\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}, c⃗=2i^−j^+2k^\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} and d⃗\vec{d} be a vector such that b⃗×d⃗=c⃗×d⃗\vec{b} \times \vec{d} = \vec{c} \times \vec{d} and a⃗⋅d⃗=4\vec{a} \cdot \vec{d} = 4. Then ∣a⃗×d⃗∣2|\vec{a} \times \vec{d}|^2 is equal to _____

Answer

Correct answer:128

Step-by-step solution

Standard Method

Given: a⃗=i^+2j^+k^\vec{a} = \hat{i} + 2\hat{j} + \hat{k}, b⃗=3i^−3j^+3k^\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}, c⃗=2i^−j^+2k^\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}, b⃗×d⃗=c⃗×d⃗\vec{b} \times \vec{d} = \vec{c} \times \vec{d} and a⃗⋅d⃗=4\vec{a} \cdot \vec{d} = 4.

Find: ∣a⃗×d⃗∣2|\vec{a} \times \vec{d}|^2.

From

b⃗×d⃗−c⃗×d⃗=0⃗\vec{b} \times \vec{d} - \vec{c} \times \vec{d} = \vec{0}

we get

(b⃗−c⃗)×d⃗=0⃗(\vec{b} - \vec{c}) \times \vec{d} = \vec{0}

So, d⃗\vec{d} is parallel to b⃗−c⃗\vec{b} - \vec{c}, hence

d⃗=λ(b⃗−c⃗)\vec{d} = \lambda (\vec{b} - \vec{c})

for some scalar λ\lambda.

Now,

b⃗−c⃗=(3−2)i^+(−3−(−1))j^+(3−2)k^=i^−2j^+k^\vec{b} - \vec{c} = (3 - 2)\hat{i} + (-3 - (-1))\hat{j} + (3 - 2)\hat{k} = \hat{i} - 2\hat{j} + \hat{k}

Therefore,

d⃗=λ(i^−2j^+k^)\vec{d} = \lambda (\hat{i} - 2\hat{j} + \hat{k})

Using a⃗⋅d⃗=4\vec{a} \cdot \vec{d} = 4,

(i^+2j^+k^)⋅(λi^−2λj^+λk^)=4(\hat{i} + 2\hat{j} + \hat{k}) \cdot (\lambda \hat{i} - 2\lambda \hat{j} + \lambda \hat{k}) = 4 λ−4λ+λ=4\lambda - 4\lambda + \lambda = 4 −2λ=4-2\lambda = 4 λ=−2\lambda = -2

Hence,

d⃗=−2i^+4j^−2k^\vec{d} = -2\hat{i} + 4\hat{j} - 2\hat{k}

Now compute the cross product:

a⃗×d⃗=∣i^j^k^121−24−2∣\vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix} =i^(2(−2)−1(4))−j^(1(−2)−1(−2))+k^(1(4)−2(−2))= \hat{i}(2(-2) - 1(4)) - \hat{j}(1(-2) - 1(-2)) + \hat{k}(1(4) - 2(-2)) =−8i^+8k^= -8\hat{i} + 8\hat{k}

Therefore,

∣a⃗×d⃗∣2=(−8)2+02+82=64+64=128|\vec{a} \times \vec{d}|^2 = (-8)^2 + 0^2 + 8^2 = 64 + 64 = 128

So, the value of ∣a⃗×d⃗∣2|\vec{a} \times \vec{d}|^2 is 128128.

The solution concludes with the correct answer as 128128.

Using Vector Identity

Given: a⃗⋅d⃗=4\vec{a} \cdot \vec{d} = 4 and from b⃗×d⃗=c⃗×d⃗\vec{b} \times \vec{d} = \vec{c} \times \vec{d}, we get d⃗=λ(b⃗−c⃗)\vec{d} = \lambda (\vec{b} - \vec{c}).

Find: ∣a⃗×d⃗∣2|\vec{a} \times \vec{d}|^2.

First,

b⃗−c⃗=i^−2j^+k^\vec{b} - \vec{c} = \hat{i} - 2\hat{j} + \hat{k}

So,

d⃗=λ(i^−2j^+k^)\vec{d} = \lambda (\hat{i} - 2\hat{j} + \hat{k})

Using the dot product condition,

(i^+2j^+k^)⋅λ(i^−2j^+k^)=4(\hat{i} + 2\hat{j} + \hat{k}) \cdot \lambda (\hat{i} - 2\hat{j} + \hat{k}) = 4 λ(1−4+1)=4\lambda (1 - 4 + 1) = 4 −2λ=4⇒λ=−2-2\lambda = 4 \Rightarrow \lambda = -2

Hence,

d⃗=−2i^+4j^−2k^\vec{d} = -2\hat{i} + 4\hat{j} - 2\hat{k}

Now use the identity

∣a⃗×d⃗∣2+(a⃗⋅d⃗)2=∣a⃗∣2∣d⃗∣2|\vec{a} \times \vec{d}|^2 + (\vec{a} \cdot \vec{d})^2 = |\vec{a}|^2 |\vec{d}|^2

We have

∣a⃗∣2=12+22+12=6|\vec{a}|^2 = 1^2 + 2^2 + 1^2 = 6 ∣d⃗∣2=(−2)2+42+(−2)2=24|\vec{d}|^2 = (-2)^2 + 4^2 + (-2)^2 = 24 (a⃗⋅d⃗)2=42=16(\vec{a} \cdot \vec{d})^2 = 4^2 = 16

Therefore,

∣a⃗×d⃗∣2=6×24−16=144−16=128|\vec{a} \times \vec{d}|^2 = 6 \times 24 - 16 = 144 - 16 = 128

Therefore, the required value is 128128.

Common mistakes

  • Assuming directly that d⃗=b⃗−c⃗\vec{d} = \vec{b} - \vec{c}. This is wrong because (b⃗−c⃗)×d⃗=0⃗(\vec{b} - \vec{c}) \times \vec{d} = \vec{0} only implies parallelism, not equality. Write d⃗=λ(b⃗−c⃗)\vec{d} = \lambda (\vec{b} - \vec{c}) and then use the dot product condition to find λ\lambda.

  • Making a sign error while computing b⃗−c⃗\vec{b} - \vec{c}. The j^\hat{j} component is −3−(−1)=−2-3 - (-1) = -2, not −4-4 or 22. Recompute each component carefully before substituting into d⃗=λ(b⃗−c⃗)\vec{d} = \lambda (\vec{b} - \vec{c}).

  • Using the identity ∣a⃗×d⃗∣=∣a⃗∣ ∣d⃗∣|\vec{a} \times \vec{d}| = |\vec{a}|\,|\vec{d}| without checking whether the vectors are perpendicular. This is wrong because that equality holds only for a right angle. Use either the determinant method or ∣a⃗×d⃗∣2+(a⃗⋅d⃗)2=∣a⃗∣2∣d⃗∣2|\vec{a} \times \vec{d}|^2 + (\vec{a} \cdot \vec{d})^2 = |\vec{a}|^2 |\vec{d}|^2.

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