Verified Solution • 2025–26 Session moderate

QJEEM23Ma45e4: JEE Main 2023 Physics - Class 11 JEEM23Ma45e4 Solution: Let and . If a vector satisfies and , then is equal to :

Let a⃗=i^+4j^+2k^,b⃗=3i^−2j^+7k^\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k} and c⃗=2i^−j^+4k^\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k} . If a vector d⃗\vec{d} satisfies d⃗×b⃗=c⃗×b⃗\vec{d} \times \vec{b}=\vec{c} \times \vec{b} and d⃗⋅a⃗=24\vec{d} \cdot \vec{a}=24 , then ∣d⃗∣2|\vec{d}|^{2} is equal to :

Multiple Choice Options

  • A 313
  • B 413
  • C 423
  • D 323

Step-by-Step Verified Solution

Given that d⃗×b⃗=c⃗×b⃗\vec{d} \times \vec{b} = \vec{c} \times \vec{b} , we can rewrite this as:

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

This implies that the vector d⃗−c⃗\vec{d} - \vec{c} is a scalar multiple of b⃗\vec{b} :

d⃗=c⃗+λb⃗\vec{d} = \vec{c} + \lambda \vec{b}

Also, we are given that d⃗⋅a⃗=24\vec{d} \cdot \vec{a} = 24 :

(c⃗+λb⃗)⋅a⃗=24(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24

Now, we can find the value of λ\lambda :

λ=24−a⃗⋅c⃗b⃗⋅a⃗=24−69=2\lambda = \frac{24 - \vec{a} \cdot \vec{c}}{\vec{b} \cdot \vec{a}} = \frac{24 - 6}{9} = 2

Therefore, we have :

d⃗=c⃗+2(b⃗)=8i^−5j^+18k^\vec{d} = \vec{c} + 2(\vec{b}) = 8\hat{i} - 5\hat{j} + 18\hat{k}

Now, we can find the squared magnitude of d⃗\vec{d} :

∣d⃗∣2=64+25+324=413|\vec{d}|^2 = 64 + 25 + 324 = 413

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